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

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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

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

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