Mean Geometry. Zachary R. Abel. July 9, 2007

Size: px
Start display at page:

Download "Mean Geometry. Zachary R. Abel. July 9, 2007"

Transcription

1 Mean Geometry Zachary R. bel July 9, 2007

2 ontents Introduction 3. asic efinitions oint verages Figure verages Finally! roofs of MGT 5 2. roof : Spiral Similarity roof 2: omplex Numbers Some roblems 7 3. Equilaterals Joined at the Hip Napoleon s Last Hurrah Extending MGT? ut Wait, There s More! The symmetric ropeller Figure ddition 4. When id the IM Get So Easy? Fresh Look at an ld Result Few Harder roblems 3 5. retty(,) usy iagram Hidden ircles Nagel Who? dditional roblems 9 7 Solutions to dditional roblems 2

3 List of Figures irect and inverse similarity oint averaging Figure averaging MGT for triangles Similar triangles i i Similar triangles i i roblem Napoleon s Theorem Equilateral triangle JKL More Napoleon roblem roblem roblem oint addition roblem ompeiu s Theorem roblem Studying quadrilateral uadrilateral roblem ircles and Three similar triangles roblem Excircle I a roblem roblem roblem roblem 4.c Van ubel s Theorem roblem 6.a roblem 6.b roblem 6.c roblem 6.d roof 2 of perpendicularization lemma roblem 8.b roblem 8.c Generalization of problem roblem roblem roblem

4 Introduction (for lack of a better section header) There are generally two opposite approaches to lympiad geometry. Some prefer to draw the diagram and simply stare (labeling points only clutters the diagram!), waiting for the interactions between the problem s various elements to present themselves visually. thers toss the diagram onto the complex or coordinate plane and attempt to establish the necessary connections through algebraic calculation rather than geometric insight. This article discusses an interesting way to visualize and approach a variety of geometry problems by combining these two common methods: synthetic and analytic. We ll focus on a theorem known as The Fundamental Theorem of irectly Similar Figures [4] or The Mean Geometry Theorem (abbreviated here as MGT). lthough the result is quite simple, it nevertheless encourages a powerful new point of view.. asic efinitions figure has positive orientation if its vertices are listed in counterclockwise order (like pentagons E or LMN in diagram ), and otherwise it has negative orientation (like V W Y Z or even E). Two figures are similar if..., well, we re all familiar with similar figures; two similar figures are directly similar if they have the same orientation, and otherwise they re inversely similar. For example, pentagon E is directly similar to pentagon LMN, but inversely similar to V W Y Z. E Y Z N M L W V Figure : irect and inverse similarity.2 oint verages Now for some non-standard notation. We define the average of two points and as the midpoint M of segment, and we write = M. We can also compute weighted averages with weights other than 2 and 2, as long as the weights add to : the weighted average ( k) + (k) is the point on line so that / = k. (This is consistent with the complex-number model of the plane, though we will still treat and as points rather than complex values.) For example, N = is one-third of the way across from to (see Figure 2) = M N = Figure 2: oint averaging.3 Figure verages We can use the concept of averaging points to define the (weighted) average of two figures, where a figure may be a polygon, circle, or any other 2-dimensional path or region (this article will stick to polygons). To average two figures, simply take the appropriate average of corresponding pairs of points. For polygons, we need only average corresponding vertices: 3

5 4 4 2 i + 2 i = i for i =, 2, 3, 4, so = Figure 3: Figure averaging.4 Finally! y now you can probably guess the punchline. Mean Geometry Theorem. The (weighted) average of two directly similar figures is directly similar to the two original figures. Look again at Figure 3. uadrilaterals and are directly similar, so the Mean Geometry Theorem guarantees that their average, 2 3 4, is directly similar to both of them. 4

6 2 roofs of MGT We mentioned in the introduction that this article incorporates some ideas from both synthetic and analytic geometry. Here we offer two proofs of the Theorem one based in each category to better illustrate the connection. lthough MGT holds for all types of figures, it suffices to prove result for triangles (why?): 3 3 MGT for Triangles. If a = 2 3 and b = 2 3 are directly similar triangles in the plane, then the weighted average c = ( k) (k) 2 3 = 2 3 is similar to both a and b (Figure 4) roof : Spiral Similarity If b is simply a translation of a, say by vector V, then the translation by vector k V sends a to c, so they are directly similar (in fact congruent). therwise, there exists a unique spiral similarity r θ that takes a to b (see [5, Theorem 4.82]). The three triangles i i (for i =, 2, 3) have i i = θ and i / i = r, so they are all similar (see Figure 5). Figure 4: MGT for triangles Figure 5: Similar triangles i i Figure 6: Similar triangles i i Since, 2, and 3 are in corresponding positions in these three similar triangles (since i i / i i = k for i =, 2, 3), the three triangles i i (i =, 2, 3) are similar to each other (Figure 6). Thus the spiral similarity r θ, where θ = and r = /, sends 2 3 to 2 3, so these triangles are directly similar, as desired. 2.2 roof 2: omplex Numbers We ll use capital letters for points and lower case for the corresponding complex numbers: point corresponds to the complex number a, and so on. i.e. a rotation through an angle θ followed by a dilation with ratio r, both centered at the point 5

7 Since the triangles are similar, 2 3 = 2 3 and 3 / 2 = 3 / 2. If z is the complex number with argument 2 3 and magnitude 3 / 2, we have a3 a a 2 a = z = b3 b b 2 b. r, rearranged, a 3 = ( z)a + (z)a 2 and b 3 = ( z)b + (z)b 2. ( ) To prove the theorem, we need to show that 2 3 behaves similarly (no pun intended), i.e. we need to show that c 3 = ( z)c + (z)c 2. ut since c i = ( k)a i + (k)b i for i =, 2, 3 (by definition of c ), this equality follows directly from ( ): c 3 = ( k)a 3 + (k)b 3 = ( k) ( ( z)a + (z)a 2 ) + (k) ( ( z)b + (z)b 2 ) = ( z) ( ( k)a + (k)b ) + (z) ( ( k)a2 + (k)b 2 ) = ( z)c + (z)c 2, as desired. 6

8 3 Some roblems The theorem has been proved, but there might be some lingering doubts about the usefulness of such a seemingly simple and specialized statement. 2 In this section, we ll put the Theorem to work, and we ll learn to recognize when and how MGT may be applied to a variety of problems. 3. Equilaterals Joined at the Hip roblem (Engel). and are positively oriented regular triangles with a common vertex. Show that the midpoints of,, and are vertices of a regular triangle. (Remember: positively oriented means the vertices are listed in counterclockwise order.) E F Figure 7: roblem To show that EF is equilateral, we d like to express EF as an average of two other directly similar equilateral triangles. From the diagram we see that =, = E, and = F, and putting these together yields = EF. So we re done by MGT, right? Let s make sure everything is in place. re triangles and directly similar? The problem tells us that they are both positively oriented equilateral triangles, so yes. re we indeed taking a weighted average of the two triangles? In other words, do the weights add to? f course! =. Now we can confidently apply the Mean Geometry Theorem and conclude that triangle EF must be directly similar to and, i.e. EF is equilateral. 3.2 Napoleon s Last Hurrah roblem 2 (Napoleon s Theorem). If equilateral triangles,, R are erected externally on the sides of triangle, their centers, Y, Z form an equilateral triangle (Figure 8). 2 lease pardon the unintentional alliteration. 7

9 J R Z R L Z K Y Y Figure 8: Napoleon s Theorem Figure 9: Equilateral triangle JKL The setup looks a lot like the configuration in the first problem, where two equilaterals share a common vertex. We can try to mimic this earlier configuration by considering equilateral triangle 2 + 2R = JKL. (Figure 9) Now the way to reach the target points,, Y, and Z, presents itself: is on median J, and it s 3 of the way across. This means that = 2 3 J + 3, and likewise for Y and Z. We can write Y Z = 2 3 JKL + 3, and since both JKL and are negatively oriented equilateral triangles, we re done! 3.3 Extending MGT? Let s look closer at that solution. First we averaged and R: 2 + 2R = JKL. Then, we averaged 2 this with : 3 JKL + 3 = Y Z. Momentarily indulging ourselves in some questionable manipulation, we can substitute the first equation into the second and simplify to find R + = Y Z. ( ) 3 This suggests that we may be able to generalize MGT to three figures, like so: Extended MGT. efine the weighted average of 3 points ω + ω ω 3 3 (where the weights ω, ω 2, and ω 3 are real numbers and add to ) just as we would in the complex plane. 3 Then if 2 3, 2 3, and 2 3 are directly similar triangles, their weighted average ω a ω b ω c 2 3 is also directly similar to them. This naturally extends to include figures other than triangles (as long as they re all similar to each other) or more than 3 similar figures (as long as the weights add to ). an you prove this? nyway, by this generalized version of MGT, equation ( ) alone provides a succinct, one-line proof of Napoleon s Theorem. (ool, huh?) 3.4 ut Wait, There s More! roblem 3. Napoleon isn t done with us yet: prove that triangles and Y Z have the same centroid. 3 s a notable special case, is the centroid of triangle

10 Like Napoleon s Theorem itself, this can also be done in one line. Here s the first half: Y + Z =??? ( ) 3 efore you read ahead, try to figure out how and why this proves the problem. (Jeopardy song plays... ) k, welcome back! The three red triangles are isosceles triangles, so they are all similar. If G and H denote the centroids of and Y Z respectively, then the result of the expression in equation ( ) is triangle GHG, which (by the Theorem) must also be a triangle. ut two of its vertices are at the same place! The triangle has thus degenerated into a point, so all three of its vertices are at the same place, and G = H. R Z G Y 3.5 I an t Take ny More Equilaterals! and the symmetric ropeller Figure 0: More Napoleon Let s do one more problem of a similar flavor before we move on. roblem 4. ositively oriented equilateral triangles,, and EF share a vertex. If,, and R are the midpoints of, E, and F respectively, prove that R is equilateral. roblem 5 (rux Mathematicorum). In quadrilateral, M is the midpoint of, and three equilateral triangles E, F, and G are constructed externally. If N is the midpoint of EF and is the midpoint of F G, prove that MN is equilateral. roblem 6. The four triangles, b c, a c, and a b are directly similar, and M a, M b, and M c are the midpoints of a a, b b, and c c. Show that M a M b M c is also similar to. b R F E Figure : roblem 4 G M N F Figure 2: roblem 5 E M c c a c M b M a b a Figure 3: roblem 6 Hm, didn t I just say one more problem? Indeed, all we have to do is solve problem 6; the rest come free. The diagrams are drawn to illustrate illustrate how the first two problems are special cases of problem 6. 4 So, let s solve problem 6, known as the symmetric ropeller [2]. The problem gives us a plethora of similar triangles to work with, so our first instinct should be to try to write triangle M a M b M c as an average of these: M a M b M c = ω + ω a b c + ω b a c + ω c a b. ( ) F. 4 In problem 4, the middle triangle has degenerated into point. In problem 5, one of the outside triangles degenerates into point 9

11 The three triangles b c, a c, and a b play identical roles in the problem, so we have every reason to guess that ω a = ω b = ω c. Now, for ( ) to work, we need M = 2 ( a + a ) = (ω + ω a ) + ω a ( a + a ). If we set ω a = 2 and ω = ω a = 2, then this works out perfectly (and likewise for M b and M c ). So does M a M b M c = b c + 2 a c + 2 a b finish the proof? 5 The triangles are all similar and the weights add to (= ), so yes. Sweet, another one liner. 5 Remember that negative weights are allowed, as long as the weights add to. 0

12 4 Figure ddition I keep stressing that the weights in a weighted average must add to. What would happen if they didn t? In the complex-number proof of MGT from section 2.2, there s really no reason the weights must be k and ( k). They could be any real numbers, and the proof works the same! There is one subtle difference, though. In the diagram to the right, use weights of ω = ω 2 =, and we ll try to calculate +. Where is it? To add vectors or complex numbers like this, we need to know where the origin is. M If the origin is at, say, M, the midpoint between and, then + represents the sum of the blue vectors M + M = 0 = M. ut if we put the origin at a different point, then the sum + is now the sum of the red vectors + = =. So, with point addition, the sum depends on the location of the origin, i.e. we must first specify an origin. With this minor change, MGT Figure 4: oint addition extends yet again: MGT: Figure ddition. For real numbers ω and ω 2, define the sum of points ω + ω 2 2 as the endpoint of the vector ω + ω 2 2, where is a specified origin. Then if 2 3 and 2 3 are directly similar, the sum of figures ω a ω b 2 3 (formed by adding corresponding vertices) is directly similar to the original two triangles. s before, this naturally extends to more complicated figures and to more than two figures. There is another way to visualize figure addition through dilation. For two similar triangles and 2 2 2, the sum may be constructed by averaging the two triangles, , and then dilating this with center and ratio When id the IM Get So Easy? (answer: 977) roblem 7 (IM jury 977 [6]). and are regular triangles of the same orientation, S is the centroid of, and M and N are the midpoints of and, respectively. Show that SM SN (Figure 5). (In the diagram, points W,, Y, and Z are midpoints.) First of all, the red triangles both look like triangles, and if we can prove that, we re done. The two equilateral triangles give us plenty of s to work with; all we have to do is find the right ones! oints, Z, N, and W form a parallelogram, so if is the origin, Z + W = N. With a little experimentation, we arrive at NS = Z + W S, and since both Z and W S are positively oriented triangles, so is NS. Similarly, MS = Y +S (again with as origin) proves that MS is a negatively oriented triangle. Whew, that was quick! W N S Z Y M 4.2 Fresh Look at an ld Result (936, to be precise) Figure 5: roblem 7 roblem 8 (ompeiu s Theorem []). Given an equilateral triangle and a point that does not lie on the circumcircle of, one can construct a triangle of side lengths equal to,, and. If lies on the circumcircle, then one of these three lengths is equal to the sum of the other two. Erect equilateral triangles Y and with the same orientation as. With as origin, consider equilateral triangle Y + =. It must be equilateral with the same orientation as, which

13 Y Figure 6: ompeiu s Theorem means = Y + =, i.e. Y is a parallelogram. Notice that Y has =, Y =, and Y = =, so if it is not degenerate, Y is the triangle we re looking for. When is this triangle degenerate, i.e. when are, and Y collinear? This happens if and only if (using directed angles modulo 80 ) = Y = 60 =, i.e. quadrilateral is cyclic. nd certainly, if Y is degenerate, then one of its sides equals the sum of the other two. 2

14 5 Few Harder roblems t this point we ve become proficient in utilizing the various forms of the Theorem, and we ve learned to look for a MGT approach when the problem hands us loads of similar triangles with which to play. ut MGT can be useful in many other situations as well, even if its application may be far from obvious. In this section, we ll look at a few harder problems to stress that the MGT may not solve every problem immediately, but it s a valuable method to keep in mind as you explore a problem. It may be that MGT is only one of many steps in your solution. r, ideas related to MGT may lead you to a solution that doesn t use it at all. The point is that looking at a diagram from an MGT viewpoint may tell you things that you formerly wouldn t have noticed. This brings up another point: in order to recognize uses for MGT, you must have an accurate diagram or two, or three to look at. This article is filled with diagrams for exactly that purpose. (Whether you re solving a geometry problem using MGT or not, it s usually a good idea to have a decent diagram handy!) n that note, let s bring on the problems. 5. retty(,) usy iagram roblem 9 (IM Shortlist 2000, G6). Let be a convex quadrilateral with not parallel to, and let be a point inside such that = < 90 and = < 90. If Y is the point of intersection of the perpendicular bisectors of and, prove that Y = 2. Y Figure 7: roblem 9 First thing to notice: by simply relabeling the diagram, it must also be true that Y = 2. So, designate = α and = δ. Next, notice that triangles and are similar. oh, that means we should form another similar triangle 2 + 2! That s a decent thought, but unfortunately, even though and are similar, they re not directly similar. It may not clear where to go from here, but since it s necessary to start somewhere, we ll begin with the only tangible fact we have: similar triangles and. It s interesting that no matter how these two triangles are hinged or scaled around, point Y still has its curious property. This suggests a possible direction for exploration: scale the triangles and see what happens to point Y. Let s leave triangle fixed while we enlarge and shrink. What size triangle would be easier to analyze? How about zero! onsider quadrilateral, where has shrunken to the degenerate triangle at point (Figure 8). The corresponding Y is the intersection of the perpendicular bisectors of = and =, i.e. the circumcenter of triangle. Is it true that 2 =? Yes, since arc  of circle has measure 2α. So when triangle shrinks to zero, everything works out as expected. What other size for triangle might work well? Let s look at quadrilateral, where is congruent to as shown in Figure 9. To locate Y, we should be looking for the perpendicular bisectors 3

15 = = Y Figure 8: Studying quadrilateral of and. ut these are the same line MN, so where is Y? We can instead locate Y by finding the point on this line so that Y = 2α, since this is another property our Y s should have. This means that Y = α =, so Y is cyclic, i.e. Y is the second intersection of circle with line MN. (If the circle happens to be tangent to line M N, second intersection simply means tangency point.) M Y Y N Figure 9: uadrilateral It seems that we ve lost sight of our original problem. We ve studied quadrilaterals and, but not the original. Luckily, a perusal of diagram 9 reveals the next step: triangles, Y, and Y are similar since they re all isosceles with vertex angle 2α, so we should be able to average them. Indeed, if / = k, then we should have ( k) + (k)y = Y. This is (almost) the last step to a complete solution! full, self-contained solution is given below. Full Solution. Let / = k, and dilate and around with ratio /k to points and respectively, so that is an isosceles trapezoid. M and N are the midpoints of and. efine as the circumcenter of triangle, and let Y be the second intersection of circle with line MN. If = α and = δ, it follows that = 2α, Y = α which implies Y = 2α, and likewise, = 2δ and Y = 2δ. Therefore, the isosceles triangles and Y are similar, as are triangles and Y. efine Y = ( k) + (k)y, and notice that triangles Y = ( k) + (k)y and Y = 4

16 ( k) + (k)y are isosceles by MGT. Thus, Y is the intersection of the perpendicular bisectors of and, i.e. Y = Y. Furthermore, by our application of MGT, Y = = 2α,. 5.2 Hidden ircles roblem 0 (US TST 2005 #6). Let be an acute scalene triangle with as its circumcenter. oint lies inside triangle with = and =. oint lies on line with =. rove that = 2. M Figure 20: roblem 0 Most of the diagram is straightforward: is the circumcenter, is the intersection of with the perpendicular bisector of, and I ve added M, the intersection of with. Everything is simple to navigate except itself, so that s where we ll start investigating. The first strange angle equality, i.e. =, shows that M M. Thus, M/M = M/M, or (M)(M ) = (M) 2. This shows, by power of a point, that the circle through,, and is tangent to line M at. 6 Likewise, the circle through,, and is tangent to M at. (all these two circles [and their centers] and 2 respectively.) Finally, (M) 2 = (M)(M ) = (M) 2, so M is the midpoint of. 2 M Figure 2: ircles and 2 The diagram becomes much clearer from the viewpoint of the two circles. The circles intersect at and, and line is their common tangent. oint, being the intersection of their axis of symmetry (line 2 ) with a common tangent (line ), must be the center of homothecy between circles and 2. Let be the midpoint of. We re asked to prove that = 2, or equivalently, = M. ut triangles and M are both right triangles, so we need to prove that they are similar. Here s where 6 If this circle intersected line M at some other point, then power of a point would show that (M) 2 = (M)(M ) = (M)(M ), so M = M and =. 5

17 Mean Geometry might come in handy: to prove that these triangles are similar, we might be able to find a third triangle similar to, say, and then express triangle M as a weighted average of those two. So now we hunt for a third similar triangle. In order for this to work, this triangle should have one vertex at, one vertex along line M, and one vertex along line. What key points are on these two lines? Just and. nd since is already being used, let s think about. If triangles and M were similar to triangle Z, where would this mystery point Z have to be? We already want it on line, and since = = Z, we would also need Z to be on line. So define Z =, and let s see if Z is indeed the triangle we re looking for. First of all, is it a right triangle? ngle Z is a right angle if and only if Z is right, so we need, i.e. should be tangent to circle at. It turns out to be true, as follows. Let T be the second intersection of with circle. If r (where r = 2 / ) is the homothecy taking circle to circle 2, we have r (T ) = and r () =, so = T = T. Thus, triangles and are similar, so ()() = () 2, and is tangent to circle. So (by tracing backwards through a few lines of reasoning above), triangle Z actually does have a right angle at. This means it is similar to triangle. 2 Z T M Figure 22: Three similar triangles The next part of our initial plan was to find a weighted average of triangles Z and that would produce triangle M, thus completing the proof. So, all we need to show is that /Z = M/ M. Good luck. Those segment lengths aren t easy to calculate, even with plenty of paper and tons of time. For the first time in this article, MGT fails to miraculously save the day. ut we ve come far enough with the MGT idea so that the proof is moments away. Let r θ be the spiral dilation centered at that sends to Z. Since M is on line, r θ (M) = M is on line Z. lso, since triangle M M is similar to triangle, we have MM = 90, so M is also on the perpendicular bisector of. This means r θ (M) = M =. Thus, M =, as desired. (Notice that, even though MGT was a driving force for most of the solution, not a single mention of it is necessary in the final writeup.) 5.3 Nagel Who? roblem (USM 200 #2). Let be a triangle and let ω be its incircle. enote by and E the points where ω is tangent to sides and, respectively. enote by 2 and E 2 the points on sides and, respectively, such that 2 = and E 2 = E, and denote by the point of intersection of segments 2 and E 2. ircle ω intersects segment 2 at two points, the closer of which to the vertex is denoted by. rove that = 2. In the diagram I ve added F and F 2 to preserve symmetry, and I included the midpoints of the sides of the triangle. 6

18 E F F F 2 I E E S E 2 F I 2 R I a 2 Figure 23: roblem Figure 24: Excircle I a Using the standard notations a =, b =, c =, and s = 2 (a + b + c), it is relatively well known that = s b. Indeed, if E = F = x, F = = y, = E = z, then the system y + z = a, z + x = b, x + y = c can be solved to give = y = (c + a b)/2 = s b. Likewise, if the excircle opposite is tangent to at V, a similar calculation shows that V = s b =, i.e. V = 2. So 2, E 2, F 2 are the tangency points of,, with the triangle s three excircles. nother relatively well-known point in the diagram is, commonly referred to as the Nagel point of triangle. It s simply the intersection of the three cevians 2, E 2, and F 2, which must concur by eva s theorem: 2 2 E 2 E 2 F 2 F 2 = s c s b s a s c s b s a =. useful property of the Nagel point, 7 other than its mere existence, is how its defining cevians interact with the incircle: line 2 intersects the incircle at point diametrically opposite from. To prove this, let r = I/I a (in Figure 24), so that r is the homothecy centered at taking excircle I to incircle I. Since 2 I is perpendicular to, its image through r, namely I, is also perpendicular to. Therefore, is a diameter of circle I. The same goes for diameters E R and F S. Now to the problem. The midpoints, E, and F inspire us to consider 2 E F + 2 2E 2 F 2 = EF. ( ) Triangles E F and 2 E 2 F 2 aren t similar, so MGT doesn t tell us anything directly. ut it s still worth noticing. If it s true that = 2, it must also happen that R = E 2 and S = F 2. This means that, with as origin, we d like to be able to show that + 2 E 2 F 2 = RS. gain, these triangles are not similar, so MGT isn t applicable. ut this equation has striking similarities with ( ). These similarities become more pronounced if we rewrite equation ( ) as 2EF + 2 E 2 F 2 = E F ( ) 7 The Nagel point is also (and less commonly) known as the bisected perimeter point [8] or the splitting center [7], since the cevians 2, etc., bisect the perimeter of the triangle, i.e. + 2 = + 2 = s. 7

19 F 2 E F E F E 2 I S R 2 Figure 25: roblem Triangle 2EF is simply a translation of, and triangle E F is a translation of RS. The two equations are almost identical! This inspires the following observation: Non-similar Figure ddition. If three triangles (figures) satisfy ω + ω 2 2 = 3, and if is a translated version of, then ω + ω 2 2 is a translation of 3, regardless of the location of the origin. Indeed, if = + V, 8 then ω + ω 2 2 = ω + ω ω V = 3 + ω V. Now we can finish the problem. eginning with equation ( ), translate 2EF to coincide with. The above observation (with our origin still at ) proves that R S = + 2 E 2 F 2 must be a translation of RS. ut since + 2 = lies on 2, and likewise for R and S, triangles RS and R S are also homothetic with center. So the two triangles must be identical, and in particular, = = + 2. So = 2, proving the desired result. 8 i.e. a translation by vector V 8

20 6 dditional roblems roblem 2. Recall problem 4: ositively oriented equilateral triangles,, and EF share a vertex. If,, and R are the midpoints of, E, and F respectively, prove that R is equilateral. rove this problem using Napoleon s Theorem. roblem 3 (rux Mathematicorum). line parallel to the side of equilateral intersects at M and at, thus making M equilateral as well. is the center of M and E is the midpoint of. etermine the angles of E. roblem 4. The following theorem appears in Geometry Revisited [5] as a special case of a theorem of etersen and Schoute: If and are two directly similar triangles, while,, are three directly similar triangles, then is directly similar to. a. Show that this theorem generalizes the triangle version of MGT. b. What minor adjustment can be made to the statement of MGT to account for this generalization (and it s proof)? c. Show that Napoleon s theorem is a special case of this theorem. roblem 5 (Van ubel s Theorem). Given an arbitrary planar quadrilateral, place a square outwardly on each side, and connect the centers of opposite squares. Then these two lines perpendicular and of equal length. roblem 6. Equilateral triangles E, F, G, H are erected outwardly on the sides of a plane quadrilateral. a. Let M, N,, and be the midpoints of segments EG, HF,, and respectively. What is the shape of MN? b. M d and M a are the centroids of H and E, and equilateral triangle M d T M a is oppositely oriented with respect to. Find the angles of triangle F T G. c. M a and M c are the centroids of E and G. rove that segments M a M c and F H are perpendicular, and, in addition, F H = 3 M a M c. d. Equilateral triangles EW F, F G, GY H, and HZE are oppositely oriented with respect to. rove that quadrilaterals and W Y Z have the same area. roblem 7. Let l(, R) denote the line through point perpendicular to line R. Say that Y Z perpendicularizes if l(, ), l(y, ), and l(z, ) concur at a point. If Y Z perpendicularizes and EF, prove that Y Z also perpendicularizes any linear combination of and EF. roblem 8 (lex Zhai). a. In triangle,, E, and F are altitudes. c and b are the projections of onto and, respectively, and points E a, E c, F a, and F b are defined similarly. rove that quadrilaterals c b, E a E c, and F b F a are cyclic. b. Let be the center of circle c b, and let T a be the midpoint of altitude. Similarly define b, c, T b, and T c. If is the circumcenter of triangle, show that a T a is a parallelogram, as well as b T b and c T c. c. rove that lines a T a, b T b, and c T c are concurrent. 9

Three Lemmas in Geometry

Three Lemmas in Geometry Winter amp 2010 Three Lemmas in Geometry Yufei Zhao Three Lemmas in Geometry Yufei Zhao Massachusetts Institute of Technology yufei.zhao@gmail.com 1 iameter of incircle T Lemma 1. Let the incircle of triangle

More information

IMO Training 2008 Circles Yufei Zhao. Circles. Yufei Zhao.

IMO Training 2008 Circles Yufei Zhao. Circles. Yufei Zhao. ircles Yufei Zhao yufeiz@mit.edu 1 Warm up problems 1. Let and be two segments, and let lines and meet at X. Let the circumcircles of X and X meet again at O. Prove that triangles O and O are similar.

More information

Geometry Chapter 1 Vocabulary. coordinate - The real number that corresponds to a point on a line.

Geometry Chapter 1 Vocabulary. coordinate - The real number that corresponds to a point on a line. Chapter 1 Vocabulary coordinate - The real number that corresponds to a point on a line. point - Has no dimension. It is usually represented by a small dot. bisect - To divide into two congruent parts.

More information

Advanced Euclidean Geometry

Advanced Euclidean Geometry dvanced Euclidean Geometry What is the center of a triangle? ut what if the triangle is not equilateral?? Circumcenter Equally far from the vertices? P P Points are on the perpendicular bisector of a line

More information

Chapter 1: Essentials of Geometry

Chapter 1: Essentials of Geometry Section Section Title 1.1 Identify Points, Lines, and Planes 1.2 Use Segments and Congruence 1.3 Use Midpoint and Distance Formulas Chapter 1: Essentials of Geometry Learning Targets I Can 1. Identify,

More information

CHAPTER 1 CEVA S THEOREM AND MENELAUS S THEOREM

CHAPTER 1 CEVA S THEOREM AND MENELAUS S THEOREM HTR 1 V S THOR N NLUS S THOR The purpose of this chapter is to develop a few results that may be used in later chapters. We will begin with a simple but useful theorem concerning the area ratio of two

More information

GEOMETRY CONCEPT MAP. Suggested Sequence:

GEOMETRY CONCEPT MAP. Suggested Sequence: CONCEPT MAP GEOMETRY August 2011 Suggested Sequence: 1. Tools of Geometry 2. Reasoning and Proof 3. Parallel and Perpendicular Lines 4. Congruent Triangles 5. Relationships Within Triangles 6. Polygons

More information

Centroid: The point of intersection of the three medians of a triangle. Centroid

Centroid: The point of intersection of the three medians of a triangle. Centroid Vocabulary Words Acute Triangles: A triangle with all acute angles. Examples 80 50 50 Angle: A figure formed by two noncollinear rays that have a common endpoint and are not opposite rays. Angle Bisector:

More information

Lemmas in Euclidean Geometry 1

Lemmas in Euclidean Geometry 1 Lemmas in uclidean Geometry 1 Yufei Zhao yufeiz@mit.edu 1. onstruction of the symmedian. Let be a triangle and Γ its circumcircle. Let the tangent to Γ at and meet at. Then coincides with a symmedian of.

More information

Selected practice exam solutions (part 5, item 2) (MAT 360)

Selected practice exam solutions (part 5, item 2) (MAT 360) Selected practice exam solutions (part 5, item ) (MAT 360) Harder 8,91,9,94(smaller should be replaced by greater )95,103,109,140,160,(178,179,180,181 this is really one problem),188,193,194,195 8. On

More information

11 th Annual Harvard-MIT Mathematics Tournament

11 th Annual Harvard-MIT Mathematics Tournament 11 th nnual Harvard-MIT Mathematics Tournament Saturday February 008 Individual Round: Geometry Test 1. [] How many different values can take, where,, are distinct vertices of a cube? nswer: 5. In a unit

More information

Angles that are between parallel lines, but on opposite sides of a transversal.

Angles that are between parallel lines, but on opposite sides of a transversal. GLOSSARY Appendix A Appendix A: Glossary Acute Angle An angle that measures less than 90. Acute Triangle Alternate Angles A triangle that has three acute angles. Angles that are between parallel lines,

More information

Geometry: Euclidean. Through a given external point there is at most one line parallel to a

Geometry: Euclidean. Through a given external point there is at most one line parallel to a Geometry: Euclidean MATH 3120, Spring 2016 The proofs of theorems below can be proven using the SMSG postulates and the neutral geometry theorems provided in the previous section. In the SMSG axiom list,

More information

Factoring Patterns in the Gaussian Plane

Factoring Patterns in the Gaussian Plane Factoring Patterns in the Gaussian Plane Steve Phelps Introduction This paper describes discoveries made at the Park City Mathematics Institute, 00, as well as some proofs. Before the summer I understood

More information

Chapters 6 and 7 Notes: Circles, Locus and Concurrence

Chapters 6 and 7 Notes: Circles, Locus and Concurrence Chapters 6 and 7 Notes: Circles, Locus and Concurrence IMPORTANT TERMS AND DEFINITIONS A circle is the set of all points in a plane that are at a fixed distance from a given point known as the center of

More information

Geometry Course Summary Department: Math. Semester 1

Geometry Course Summary Department: Math. Semester 1 Geometry Course Summary Department: Math Semester 1 Learning Objective #1 Geometry Basics Targets to Meet Learning Objective #1 Use inductive reasoning to make conclusions about mathematical patterns Give

More information

Chapter 6 Notes: Circles

Chapter 6 Notes: Circles Chapter 6 Notes: Circles IMPORTANT TERMS AND DEFINITIONS A circle is the set of all points in a plane that are at a fixed distance from a given point known as the center of the circle. Any line segment

More information

Solutions to Practice Problems

Solutions to Practice Problems Higher Geometry Final Exam Tues Dec 11, 5-7:30 pm Practice Problems (1) Know the following definitions, statements of theorems, properties from the notes: congruent, triangle, quadrilateral, isosceles

More information

GEOMETRY 101* EVERYTHING YOU NEED TO KNOW ABOUT GEOMETRY TO PASS THE GHSGT!

GEOMETRY 101* EVERYTHING YOU NEED TO KNOW ABOUT GEOMETRY TO PASS THE GHSGT! GEOMETRY 101* EVERYTHING YOU NEED TO KNOW ABOUT GEOMETRY TO PASS THE GHSGT! FINDING THE DISTANCE BETWEEN TWO POINTS DISTANCE FORMULA- (x₂-x₁)²+(y₂-y₁)² Find the distance between the points ( -3,2) and

More information

Geometry in a Nutshell

Geometry in a Nutshell Geometry in a Nutshell Henry Liu, 26 November 2007 This short handout is a list of some of the very basic ideas and results in pure geometry. Draw your own diagrams with a pencil, ruler and compass where

More information

DEFINITIONS. Perpendicular Two lines are called perpendicular if they form a right angle.

DEFINITIONS. Perpendicular Two lines are called perpendicular if they form a right angle. DEFINITIONS Degree A degree is the 1 th part of a straight angle. 180 Right Angle A 90 angle is called a right angle. Perpendicular Two lines are called perpendicular if they form a right angle. Congruent

More information

GEOMETRY. Constructions OBJECTIVE #: G.CO.12

GEOMETRY. Constructions OBJECTIVE #: G.CO.12 GEOMETRY Constructions OBJECTIVE #: G.CO.12 OBJECTIVE Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic

More information

Conjectures. Chapter 2. Chapter 3

Conjectures. Chapter 2. Chapter 3 Conjectures Chapter 2 C-1 Linear Pair Conjecture If two angles form a linear pair, then the measures of the angles add up to 180. (Lesson 2.5) C-2 Vertical Angles Conjecture If two angles are vertical

More information

Elementary triangle geometry

Elementary triangle geometry Elementary triangle geometry Dennis Westra March 26, 2010 bstract In this short note we discuss some fundamental properties of triangles up to the construction of the Euler line. ontents ngle bisectors

More information

Conjectures for Geometry for Math 70 By I. L. Tse

Conjectures for Geometry for Math 70 By I. L. Tse Conjectures for Geometry for Math 70 By I. L. Tse Chapter Conjectures 1. Linear Pair Conjecture: If two angles form a linear pair, then the measure of the angles add up to 180. Vertical Angle Conjecture:

More information

Duplicating Segments and Angles

Duplicating Segments and Angles CONDENSED LESSON 3.1 Duplicating Segments and ngles In this lesson, you Learn what it means to create a geometric construction Duplicate a segment by using a straightedge and a compass and by using patty

More information

1. Determine all real numbers a, b, c, d that satisfy the following system of equations.

1. Determine all real numbers a, b, c, d that satisfy the following system of equations. altic Way 1999 Reykjavík, November 6, 1999 Problems 1. etermine all real numbers a, b, c, d that satisfy the following system of equations. abc + ab + bc + ca + a + b + c = 1 bcd + bc + cd + db + b + c

More information

Definitions, Postulates and Theorems

Definitions, Postulates and Theorems Definitions, s and s Name: Definitions Complementary Angles Two angles whose measures have a sum of 90 o Supplementary Angles Two angles whose measures have a sum of 180 o A statement that can be proven

More information

Conic Construction of a Triangle from the Feet of Its Angle Bisectors

Conic Construction of a Triangle from the Feet of Its Angle Bisectors onic onstruction of a Triangle from the Feet of Its ngle isectors Paul Yiu bstract. We study an extension of the problem of construction of a triangle from the feet of its internal angle bisectors. Given

More information

New York State Student Learning Objective: Regents Geometry

New York State Student Learning Objective: Regents Geometry New York State Student Learning Objective: Regents Geometry All SLOs MUST include the following basic components: Population These are the students assigned to the course section(s) in this SLO all students

More information

Curriculum Map by Block Geometry Mapping for Math Block Testing 2007-2008. August 20 to August 24 Review concepts from previous grades.

Curriculum Map by Block Geometry Mapping for Math Block Testing 2007-2008. August 20 to August 24 Review concepts from previous grades. Curriculum Map by Geometry Mapping for Math Testing 2007-2008 Pre- s 1 August 20 to August 24 Review concepts from previous grades. August 27 to September 28 (Assessment to be completed by September 28)

More information

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 16, 2012 8:30 to 11:30 a.m.

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 16, 2012 8:30 to 11:30 a.m. GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, August 16, 2012 8:30 to 11:30 a.m., only Student Name: School Name: Print your name and the name of your

More information

V GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN

V GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN V GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN Final round. First day. 8 form. Solutions. 1. (.linkov, Y.linkov) Minor base of trapezoid D is equal to side, and diagonal is equal to base D. The line passing

More information

0810ge. Geometry Regents Exam 0810

0810ge. Geometry Regents Exam 0810 0810ge 1 In the diagram below, ABC XYZ. 3 In the diagram below, the vertices of DEF are the midpoints of the sides of equilateral triangle ABC, and the perimeter of ABC is 36 cm. Which two statements identify

More information

Geometry Honors: Circles, Coordinates, and Construction Semester 2, Unit 4: Activity 24

Geometry Honors: Circles, Coordinates, and Construction Semester 2, Unit 4: Activity 24 Geometry Honors: Circles, Coordinates, and Construction Semester 2, Unit 4: ctivity 24 esources: Springoard- Geometry Unit Overview In this unit, students will study formal definitions of basic figures,

More information

Geometry Module 4 Unit 2 Practice Exam

Geometry Module 4 Unit 2 Practice Exam Name: Class: Date: ID: A Geometry Module 4 Unit 2 Practice Exam Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which diagram shows the most useful positioning

More information

Geometry Enduring Understandings Students will understand 1. that all circles are similar.

Geometry Enduring Understandings Students will understand 1. that all circles are similar. High School - Circles Essential Questions: 1. Why are geometry and geometric figures relevant and important? 2. How can geometric ideas be communicated using a variety of representations? ******(i.e maps,

More information

Lesson 3.1 Duplicating Segments and Angles

Lesson 3.1 Duplicating Segments and Angles Lesson 3.1 Duplicating Segments and ngles In Exercises 1 3, use the segments and angles below. Q R S 1. Using only a compass and straightedge, duplicate each segment and angle. There is an arc in each

More information

Geometry Progress Ladder

Geometry Progress Ladder Geometry Progress Ladder Maths Makes Sense Foundation End-of-year objectives page 2 Maths Makes Sense 1 2 End-of-block objectives page 3 Maths Makes Sense 3 4 End-of-block objectives page 4 Maths Makes

More information

Projective Geometry - Part 2

Projective Geometry - Part 2 Projective Geometry - Part 2 Alexander Remorov alexanderrem@gmail.com Review Four collinear points A, B, C, D form a harmonic bundle (A, C; B, D) when CA : DA CB DB = 1. A pencil P (A, B, C, D) is the

More information

Final Review Geometry A Fall Semester

Final Review Geometry A Fall Semester Final Review Geometry Fall Semester Multiple Response Identify one or more choices that best complete the statement or answer the question. 1. Which graph shows a triangle and its reflection image over

More information

11.3 Curves, Polygons and Symmetry

11.3 Curves, Polygons and Symmetry 11.3 Curves, Polygons and Symmetry Polygons Simple Definition A shape is simple if it doesn t cross itself, except maybe at the endpoints. Closed Definition A shape is closed if the endpoints meet. Polygon

More information

2. If C is the midpoint of AB and B is the midpoint of AE, can you say that the measure of AC is 1/4 the measure of AE?

2. If C is the midpoint of AB and B is the midpoint of AE, can you say that the measure of AC is 1/4 the measure of AE? MATH 206 - Midterm Exam 2 Practice Exam Solutions 1. Show two rays in the same plane that intersect at more than one point. Rays AB and BA intersect at all points from A to B. 2. If C is the midpoint of

More information

Triangle Centers MOP 2007, Black Group

Triangle Centers MOP 2007, Black Group Triangle Centers MOP 2007, Black Group Zachary Abel June 21, 2007 1 A Few Useful Centers 1.1 Symmedian / Lemmoine Point The Symmedian point K is defined as the isogonal conjugate of the centroid G. Problem

More information

Geometry. Higher Mathematics Courses 69. Geometry

Geometry. Higher Mathematics Courses 69. Geometry The fundamental purpose of the course is to formalize and extend students geometric experiences from the middle grades. This course includes standards from the conceptual categories of and Statistics and

More information

Conjunction is true when both parts of the statement are true. (p is true, q is true. p^q is true)

Conjunction is true when both parts of the statement are true. (p is true, q is true. p^q is true) Mathematical Sentence - a sentence that states a fact or complete idea Open sentence contains a variable Closed sentence can be judged either true or false Truth value true/false Negation not (~) * Statement

More information

Situation: Proving Quadrilaterals in the Coordinate Plane

Situation: Proving Quadrilaterals in the Coordinate Plane Situation: Proving Quadrilaterals in the Coordinate Plane 1 Prepared at the University of Georgia EMAT 6500 Date Last Revised: 07/31/013 Michael Ferra Prompt A teacher in a high school Coordinate Algebra

More information

GEOMETRY COMMON CORE STANDARDS

GEOMETRY COMMON CORE STANDARDS 1st Nine Weeks Experiment with transformations in the plane G-CO.1 Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point,

More information

Biggar High School Mathematics Department. National 5 Learning Intentions & Success Criteria: Assessing My Progress

Biggar High School Mathematics Department. National 5 Learning Intentions & Success Criteria: Assessing My Progress Biggar High School Mathematics Department National 5 Learning Intentions & Success Criteria: Assessing My Progress Expressions & Formulae Topic Learning Intention Success Criteria I understand this Approximation

More information

Lesson 1.1 Building Blocks of Geometry

Lesson 1.1 Building Blocks of Geometry Lesson 1.1 Building Blocks of Geometry For Exercises 1 7, complete each statement. S 3 cm. 1. The midpoint of Q is. N S Q 2. NQ. 3. nother name for NS is. 4. S is the of SQ. 5. is the midpoint of. 6. NS.

More information

1. A student followed the given steps below to complete a construction. Which type of construction is best represented by the steps given above?

1. A student followed the given steps below to complete a construction. Which type of construction is best represented by the steps given above? 1. A student followed the given steps below to complete a construction. Step 1: Place the compass on one endpoint of the line segment. Step 2: Extend the compass from the chosen endpoint so that the width

More information

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, January 26, 2012 9:15 a.m. to 12:15 p.m.

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, January 26, 2012 9:15 a.m. to 12:15 p.m. GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXMINTION GEOMETRY Thursday, January 26, 2012 9:15 a.m. to 12:15 p.m., only Student Name: School Name: Print your name and the name

More information

Projective Geometry. Stoienescu Paul. International Computer High School Of Bucharest,

Projective Geometry. Stoienescu Paul. International Computer High School Of Bucharest, Projective Geometry Stoienescu Paul International Computer High School Of Bucharest, paul98stoienescu@gmail.com Abstract. In this note, I will present some olympiad problems which can be solved using projective

More information

Cevians, Symmedians, and Excircles. MA 341 Topics in Geometry Lecture 16

Cevians, Symmedians, and Excircles. MA 341 Topics in Geometry Lecture 16 Cevians, Symmedians, and Excircles MA 341 Topics in Geometry Lecture 16 Cevian A cevian is a line segment which joins a vertex of a triangle with a point on the opposite side (or its extension). B cevian

More information

Unit 2 - Triangles. Equilateral Triangles

Unit 2 - Triangles. Equilateral Triangles Equilateral Triangles Unit 2 - Triangles Equilateral Triangles Overview: Objective: In this activity participants discover properties of equilateral triangles using properties of symmetry. TExES Mathematics

More information

Visualizing Triangle Centers Using Geogebra

Visualizing Triangle Centers Using Geogebra Visualizing Triangle Centers Using Geogebra Sanjay Gulati Shri Shankaracharya Vidyalaya, Hudco, Bhilai India http://mathematicsbhilai.blogspot.com/ sanjaybhil@gmail.com ABSTRACT. In this paper, we will

More information

7. 6 Justifying Constructions

7. 6 Justifying Constructions 31 7. 6 Justifying Constructions A Solidify Understanding Task CC BY THOR https://flic.kr/p/9qkxv Compass and straightedge constructions can be justified using such tools as: the definitions and properties

More information

Seattle Public Schools KEY to Review Questions for the Washington State Geometry End of Course Exam

Seattle Public Schools KEY to Review Questions for the Washington State Geometry End of Course Exam Seattle Public Schools KEY to Review Questions for the Washington State Geometry End of ourse Exam 1) Which term best defines the type of reasoning used below? bdul broke out in hives the last four times

More information

Inversion. Chapter 7. 7.1 Constructing The Inverse of a Point: If P is inside the circle of inversion: (See Figure 7.1)

Inversion. Chapter 7. 7.1 Constructing The Inverse of a Point: If P is inside the circle of inversion: (See Figure 7.1) Chapter 7 Inversion Goal: In this chapter we define inversion, give constructions for inverses of points both inside and outside the circle of inversion, and show how inversion could be done using Geometer

More information

GEOMETRY FINAL EXAM REVIEW

GEOMETRY FINAL EXAM REVIEW GEOMETRY FINL EXM REVIEW I. MTHING reflexive. a(b + c) = ab + ac transitive. If a = b & b = c, then a = c. symmetric. If lies between and, then + =. substitution. If a = b, then b = a. distributive E.

More information

Solving Simultaneous Equations and Matrices

Solving Simultaneous Equations and Matrices Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering

More information

Week 1 Chapter 1: Fundamentals of Geometry. Week 2 Chapter 1: Fundamentals of Geometry. Week 3 Chapter 1: Fundamentals of Geometry Chapter 1 Test

Week 1 Chapter 1: Fundamentals of Geometry. Week 2 Chapter 1: Fundamentals of Geometry. Week 3 Chapter 1: Fundamentals of Geometry Chapter 1 Test Thinkwell s Homeschool Geometry Course Lesson Plan: 34 weeks Welcome to Thinkwell s Homeschool Geometry! We re thrilled that you ve decided to make us part of your homeschool curriculum. This lesson plan

More information

ON THE SIMSON WALLACE THEOREM

ON THE SIMSON WALLACE THEOREM South Bohemia Mathematical Letters Volume 21, (2013), No. 1, 59 66. ON THE SIMSON WALLACE THEOREM PAVEL PECH 1, EMIL SKŘÍŠOVSKÝ2 Abstract. The paper deals with the well-known Simson Wallace theorem and

More information

Incenter Circumcenter

Incenter Circumcenter TRIANGLE: Centers: Incenter Incenter is the center of the inscribed circle (incircle) of the triangle, it is the point of intersection of the angle bisectors of the triangle. The radius of incircle is

More information

39 Symmetry of Plane Figures

39 Symmetry of Plane Figures 39 Symmetry of Plane Figures In this section, we are interested in the symmetric properties of plane figures. By a symmetry of a plane figure we mean a motion of the plane that moves the figure so that

More information

A summary of definitions, postulates, algebra rules, and theorems that are often used in geometry proofs:

A summary of definitions, postulates, algebra rules, and theorems that are often used in geometry proofs: summary of definitions, postulates, algebra rules, and theorems that are often used in geometry proofs: efinitions: efinition of mid-point and segment bisector M If a line intersects another line segment

More information

5.1 Midsegment Theorem and Coordinate Proof

5.1 Midsegment Theorem and Coordinate Proof 5.1 Midsegment Theorem and Coordinate Proof Obj.: Use properties of midsegments and write coordinate proofs. Key Vocabulary Midsegment of a triangle - A midsegment of a triangle is a segment that connects

More information

Area. Area Overview. Define: Area:

Area. Area Overview. Define: Area: Define: Area: Area Overview Kite: Parallelogram: Rectangle: Rhombus: Square: Trapezoid: Postulates/Theorems: Every closed region has an area. If closed figures are congruent, then their areas are equal.

More information

Teaching Mathematics Vocabulary Using Hands-On Activities From an MSP Grant Summer Institute

Teaching Mathematics Vocabulary Using Hands-On Activities From an MSP Grant Summer Institute Teaching Mathematics Vocabulary Using Hands-On Activities From an MSP Grant Summer Institute Dr. Carroll G. Wells (Co-authors: Dr. Randy Bouldin, Dr. Ben Hutchinson, Dr. Candice McQueen) Department of

More information

Senior Math Circles: Geometry I

Senior Math Circles: Geometry I Universit of Waterloo Facult of Mathematics entre for Education in Mathematics and omputing pening Problem (a) If 30 7 = + + z Senior Math ircles: Geometr I, where, and z are positive integers, then what

More information

Intermediate Math Competition series (IM), Session 2, day 1, IDEA MATH Lexington, Massachusetts 1. Special angles

Intermediate Math Competition series (IM), Session 2, day 1, IDEA MATH Lexington, Massachusetts 1. Special angles Intermediate Math ompetition series (IM), Session 2, day 1, I MTH Lexington, Massachusetts 1 Special angles Zuming Feng hillips xeter cademy zfeng@exeter.edu Sentry theorem xample 1.1. lex walks along

More information

Lesson 5-3: Concurrent Lines, Medians and Altitudes

Lesson 5-3: Concurrent Lines, Medians and Altitudes Playing with bisectors Yesterday we learned some properties of perpendicular bisectors of the sides of triangles, and of triangle angle bisectors. Today we are going to use those skills to construct special

More information

Collinearity and concurrence

Collinearity and concurrence Collinearity and concurrence Po-Shen Loh 23 June 2008 1 Warm-up 1. Let I be the incenter of ABC. Let A be the midpoint of the arc BC of the circumcircle of ABC which does not contain A. Prove that the

More information

Geometry 8-1 Angles of Polygons

Geometry 8-1 Angles of Polygons . Sum of Measures of Interior ngles Geometry 8-1 ngles of Polygons 1. Interior angles - The sum of the measures of the angles of each polygon can be found by adding the measures of the angles of a triangle.

More information

Equation of a Line. Chapter H2. The Gradient of a Line. m AB = Exercise H2 1

Equation of a Line. Chapter H2. The Gradient of a Line. m AB = Exercise H2 1 Chapter H2 Equation of a Line The Gradient of a Line The gradient of a line is simpl a measure of how steep the line is. It is defined as follows :- gradient = vertical horizontal horizontal A B vertical

More information

Parallel and Perpendicular. We show a small box in one of the angles to show that the lines are perpendicular.

Parallel and Perpendicular. We show a small box in one of the angles to show that the lines are perpendicular. CONDENSED L E S S O N. Parallel and Perpendicular In this lesson you will learn the meaning of parallel and perpendicular discover how the slopes of parallel and perpendicular lines are related use slopes

More information

The Geometry of Piles of Salt Thinking Deeply About Simple Things

The Geometry of Piles of Salt Thinking Deeply About Simple Things The Geometry of Piles of Salt Thinking Deeply About Simple Things PCMI SSTP Tuesday, July 15 th, 2008 By Troy Jones Willowcreek Middle School Important Terms (the word line may be replaced by the word

More information

Bisections and Reflections: A Geometric Investigation

Bisections and Reflections: A Geometric Investigation Bisections and Reflections: A Geometric Investigation Carrie Carden & Jessie Penley Berry College Mount Berry, GA 30149 Email: ccarden@berry.edu, jpenley@berry.edu Abstract In this paper we explore a geometric

More information

Geometry: Unit 1 Vocabulary TERM DEFINITION GEOMETRIC FIGURE. Cannot be defined by using other figures.

Geometry: Unit 1 Vocabulary TERM DEFINITION GEOMETRIC FIGURE. Cannot be defined by using other figures. Geometry: Unit 1 Vocabulary 1.1 Undefined terms Cannot be defined by using other figures. Point A specific location. It has no dimension and is represented by a dot. Line Plane A connected straight path.

More information

56 questions (multiple choice, check all that apply, and fill in the blank) The exam is worth 224 points.

56 questions (multiple choice, check all that apply, and fill in the blank) The exam is worth 224 points. 6.1.1 Review: Semester Review Study Sheet Geometry Core Sem 2 (S2495808) Semester Exam Preparation Look back at the unit quizzes and diagnostics. Use the unit quizzes and diagnostics to determine which

More information

CCGPS UNIT 3 Semester 1 ANALYTIC GEOMETRY Page 1 of 32. Circles and Volumes Name:

CCGPS UNIT 3 Semester 1 ANALYTIC GEOMETRY Page 1 of 32. Circles and Volumes Name: GPS UNIT 3 Semester 1 NLYTI GEOMETRY Page 1 of 3 ircles and Volumes Name: ate: Understand and apply theorems about circles M9-1.G..1 Prove that all circles are similar. M9-1.G.. Identify and describe relationships

More information

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name:

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name: GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, August 18, 2010 8:30 to 11:30 a.m., only Student Name: School Name: Print your name and the name of

More information

Line Segments, Rays, and Lines

Line Segments, Rays, and Lines HOME LINK Line Segments, Rays, and Lines Family Note Help your child match each name below with the correct drawing of a line, ray, or line segment. Then observe as your child uses a straightedge to draw

More information

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Tuesday, August 13, 2013 8:30 to 11:30 a.m., only.

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Tuesday, August 13, 2013 8:30 to 11:30 a.m., only. GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Tuesday, August 13, 2013 8:30 to 11:30 a.m., only Student Name: School Name: The possession or use of any communications

More information

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 13, 2015 8:30 to 11:30 a.m., only.

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 13, 2015 8:30 to 11:30 a.m., only. GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, August 13, 2015 8:30 to 11:30 a.m., only Student Name: School Name: The possession or use of any communications

More information

For the circle above, EOB is a central angle. So is DOE. arc. The (degree) measure of ù DE is the measure of DOE.

For the circle above, EOB is a central angle. So is DOE. arc. The (degree) measure of ù DE is the measure of DOE. efinition: circle is the set of all points in a plane that are equidistant from a given point called the center of the circle. We use the symbol to represent a circle. The a line segment from the center

More information

Lesson 1: Introducing Circles

Lesson 1: Introducing Circles IRLES N VOLUME Lesson 1: Introducing ircles ommon ore Georgia Performance Standards M9 12.G..1 M9 12.G..2 Essential Questions 1. Why are all circles similar? 2. What are the relationships among inscribed

More information

NAME DATE PERIOD. Study Guide and Intervention

NAME DATE PERIOD. Study Guide and Intervention opyright Glencoe/McGraw-Hill, a division of he McGraw-Hill ompanies, Inc. 5-1 M IO tudy Guide and Intervention isectors, Medians, and ltitudes erpendicular isectors and ngle isectors perpendicular bisector

More information

116 Chapter 6 Transformations and the Coordinate Plane

116 Chapter 6 Transformations and the Coordinate Plane 116 Chapter 6 Transformations and the Coordinate Plane Chapter 6-1 The Coordinates of a Point in a Plane Section Quiz [20 points] PART I Answer all questions in this part. Each correct answer will receive

More information

MA 323 Geometric Modelling Course Notes: Day 02 Model Construction Problem

MA 323 Geometric Modelling Course Notes: Day 02 Model Construction Problem MA 323 Geometric Modelling Course Notes: Day 02 Model Construction Problem David L. Finn November 30th, 2004 In the next few days, we will introduce some of the basic problems in geometric modelling, and

More information

Circle Name: Radius: Diameter: Chord: Secant:

Circle Name: Radius: Diameter: Chord: Secant: 12.1: Tangent Lines Congruent Circles: circles that have the same radius length Diagram of Examples Center of Circle: Circle Name: Radius: Diameter: Chord: Secant: Tangent to A Circle: a line in the plane

More information

SIMSON S THEOREM MARY RIEGEL

SIMSON S THEOREM MARY RIEGEL SIMSON S THEOREM MARY RIEGEL Abstract. This paper is a presentation and discussion of several proofs of Simson s Theorem. Simson s Theorem is a statement about a specific type of line as related to a given

More information

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, June 20, 2012 9:15 a.m. to 12:15 p.m., only Student Name: School Name: Print your name and the name

More information

Cabri Geometry Application User Guide

Cabri Geometry Application User Guide Cabri Geometry Application User Guide Preview of Geometry... 2 Learning the Basics... 3 Managing File Operations... 12 Setting Application Preferences... 14 Selecting and Moving Objects... 17 Deleting

More information

EXPECTED BACKGROUND KNOWLEDGE

EXPECTED BACKGROUND KNOWLEDGE MOUL - 3 oncurrent Lines 12 ONURRNT LINS You have already learnt about concurrent lines, in the lesson on lines and angles. You have also studied about triangles and some special lines, i.e., medians,

More information

Geometry Regents Review

Geometry Regents Review Name: Class: Date: Geometry Regents Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. If MNP VWX and PM is the shortest side of MNP, what is the shortest

More information

GeoGebra. 10 lessons. Gerrit Stols

GeoGebra. 10 lessons. Gerrit Stols GeoGebra in 10 lessons Gerrit Stols Acknowledgements GeoGebra is dynamic mathematics open source (free) software for learning and teaching mathematics in schools. It was developed by Markus Hohenwarter

More information

alternate interior angles

alternate interior angles alternate interior angles two non-adjacent angles that lie on the opposite sides of a transversal between two lines that the transversal intersects (a description of the location of the angles); alternate

More information

Section 1.1. Introduction to R n

Section 1.1. Introduction to R n The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to

More information

Warm Up #23: Review of Circles 1.) A central angle of a circle is an angle with its vertex at the of the circle. Example:

Warm Up #23: Review of Circles 1.) A central angle of a circle is an angle with its vertex at the of the circle. Example: Geometr hapter 12 Notes - 1 - Warm Up #23: Review of ircles 1.) central angle of a circle is an angle with its verte at the of the circle. Eample: X 80 2.) n arc is a section of a circle. Eamples:, 3.)

More information