1 Geometry. 1 Cough 2012 TSTST, Problem 4.

Transcription

1 1 Geometry Initially your stumbling block is going to be not knowing the right tools to solve a problem. This will hopefully change very quickly. Once it does, you re back to the old How would I have thought of that? I ve never found a particularly satisfying answer for this question, partially because I don t claim to understand what geometric intuition is. I hope the answer is not recognizing known configurations ; however, this turns out to be an extremely important skill in many problems 1. One of the most famous is Yufei Zhao s Cyclic Quadrilaterals The Big Picture, which can be found online at If you are comfortable with everything in the first few sections, you should read this instead. Those of you who are just setting out should take the time to learn the basic tools well first. I think I ve included the majority of them here. Anyways, here are some general tips. Draw large, in-scale diagrams, if you can afford the time. Note the pluralization! At worst, this will prevent you from trying to prove something that is false. At best, this will suggest something that is true. (This is why drawing more than one diagram is helpful.) Work both forwards and backwards. One of the really nice things about geometry is that you can often go in both directions. If you keep a list of things you know and things you want, you are done the moment something appears on both lists. Simplify the problem. In that vein, you can often work backwards and simplify the end condition. If you can eliminate entire points or entire lines, this is usually a good thing. Problem 7 from the TSTST 2012 is a really good example of this which I will show you later. Phantom points. Let s say ABC is inscribed in ω, D is the intersection of ω with some line l, and you want to prove D lies on a line l 2. It s often easier to let l 1 and l 2 intersect at D and try to prove that ABCD is cyclic. Use all the givens. Occasionally there will be a screwball problem where one of the conditions isn t necessary, but usually you do need all the conditions. If you haven t used all the conditions, then there s no way you can solve the problem. Mark information as you find it. If two important angles are equal, draw them in. If a quadrilateral is cyclic, make a note of this; I like to circle the four points rather than try and draw in a circumcircle. (And for this, I usually use different colored pens). 1 Cough 2012 TSTST, Problem 4. 12

2 1.1 Computation in Triangles Notation Let ABC be triangle. Let a, b, c denote the lengths BC, CA, AB. Let K denote the area of ABC. Let s = a+b+c 2. Let R be the radius of the circumcircle of ABC, called the circumradius, and let r bet he radius of the incircle of ABC, called the inradius. For arbitrary points X, Y, and Z, [XY Z] will denote the area of XY Z Tools These are all pretty useful things to know. They also happen to be not too difficult to prove, so I will let you attempt to do so. 1. (Extended Sine Law) Prove that by showing 2. (Areas) Prove that a sin A = b sin B = c sin C = 2R a sin A = 2R, via constructing a diameter with one endpoint at C. K = s(s a)(s b)(s c) = 1 2 ab sin C = 1 2 bc sin A = 1 ca sin B 2 = sr = abc 4R 3. (Stewart s Theorem 2 ) Let D be a point on BC. Denote n = DC, m = DB and d = AD. Prove that man + dad = bmb + cnc The above lemmas are useful for computational geometry; in particular, the second one lets you solve for r and R given a, b, c, etc., and so on. 2 A man and his dad put a bomb in the sink. 13

3 1.2 Triangle Centers Vocabulary Lesson Let ABC be a triangle. Then Definition. The orthocenter of ABC, usually denoted by H, is the intersection of the perpendiculars from A to BC, B to CA and C to AB. Definition. The centroid of ABC, usually denoted by G, is the intersection of the lines joining the midpoint of a side with the opposite vertices (i.e. the medians). Definition. The incenter of ABC, usually denoted by I, is the intersection of the angle bisectors of the angles of ABC. It is also the center of a circle tangent to all three sides, called the incircle. Definition. The circumcenter of ABC, usually denoted by O, is the center of the unique circle passing through ABC, which is called the circumcircle. In a few moments, you will prove that these points do exist Ceva s Theorem A cevian is a line joining one of the vertices of a triangle to a point on the opposite side. There is a nice criterion for when three cevians are concurrent, as follows: Theorem 2 (Ceva). In ABC, point D, E, and F lie on BC, CA and AB. Then AD, BE and CF concur if and only if BD CE AF DC EA F B = 1. A F P E B D C Figure 1.1: Ceva s Theorem. The cevians concur at a point P. Exercise. Let P be a point on the cevian AD of triangle ABC. (i) Prove that BD DC = [BP D] [DP C] = [BAD] [DAC]. (ii) Use the above to show that BD DC (iii) Use this to deduce Ceva s Theorem. = [BP A] [CP A]. 14

4 1.2.3 Problems 1. (Trig Ceva) Let AD, BE and CF be cevians of a triangle ABC. Prove that they concur if and only if sin BAD sin CBE sin ACF sin CAD sin ABE sin BCF = Let ABC be a triangle and denote by M the midpoint of BC. A point P is selected on AM. BP meets AC at J, and CP meets AB at K. Show that JK BC. 3. Prove that the centroid, orthocenter, circumcenter, and incenter all exist, using either Ceva or Trig Ceva, or otherwise. 15

5 1.3 More on Length Chasing Back to MathCounts Here are some things to remember from MathCounts geometry! because I assume you are familiar with them. I mention them briefly Similar Triangles: I m sure you know what these are. Similar triangles provide a link between ratios and angles by the AA, SAS and SSS criteria for similarity. Know them when you see them. Pythagorean Theorem: Useful on the AIME. Please tell me you know what this is. It does see occasional usage on olympiads. Law of Cosines: Generalizes the Pythagorean Theorem. Quite useful on the AIME. Tangents: The two tangents from a point to a circle are equal in length, and form right angles with the center. Know this. Angle Bisector Theorem: If the angle bisector of BAC hits BC at D, then DB DC = AB AC. Like similar triangles, this gives you a way of converting equal angles into a ratio of sides. Power of a Point: We ll go over this in more detail with cyclic quadrilaterals, but you should know the statement of this! Let ω be a circle and P be a point. Suppose l through P intersects ω at A and B; then the power of P with respect to ω is defined as P A P B, and does not depend on the line l. Usually, the power is defined as positive when P is outside the circle and negative when it is inside. These are all very useful for side-chasing Ptolemy s Theorem When dealing with cyclic quadrilaterals, Ptolemy s Theorem often provides a base for computing side lengths. Theorem 3 (Ptolemy s Theorem). Let ABCD be a cyclic quadrilateral. Then AB CD + AD BC = AC BD. Sketch of Proof. Ptolemy can actually be deduced by Stewart s Theorem (and vice-versa). Let P be the intersection of the diagonals and apply Stewart on triangle ABC with cevian BP, in the form ( ) AP P C AC P B + P B = AB2 P C + BC2 P A P B P B Then use similar triangles. Exercise. Complete the proof above! 16

6 1.3.3 Problems 1. (Mu-Alpha-Theta 1991) Given triangle ABC, compute tan C given a 3 + b 3 + c 3 a + b + c = c Show that the medians of a triangle partition it into six triangles of equal area. 3. The side lengths of a triangle are 30, 40 and 50. What is the length of the shortest altitude? 4. (Ray Li) Let ABC be a triangle. The angle bisector of A meets BC at D and the circumcircle of ABC at E. If AB = 36, BC = 40, CA = 44, compute DE 2. 17

7 1.4 Angle Chasing Angles are very useful in geometry. This section is dedicated to exploring how to use angles properly Cyclic Quadrilaterals Cyclic quadrilaterals come up EVERYWHERE. It s just plain hard to do geometry if you don t understand the basics of how circles and angles interact. Definition. A quadrilateral is said to by cyclic if it can be inscribed in a circle. Here are the main results: Theorem 4. Let ABCD be a quadrilateral. ABCD is cyclic if and only if (i) DAC = DBC (ii) ABC + ADC = 180. A B D C Figure 1.2: Bob the cyclic quadrilateral. Mark ALL the angles. You should recognize the first two statements as trivial consequences of the Inscribed Angle Theorem 3. However, it is extremely important to recognize that the converse of these statements are all true! Furthermore, we have from a few sections ago: Theorem 5 (Power of a Point). A, B, C, and D lie on a circle in some order. Let P be the intersection of lines AC and BD. Then P A P C = P B P D. This quantity is called the power of P with respect to the circle, and is negative when P is inside the circle and positive otherwise. Exercise. (This is very useful.) Suppose you are instead given P = AC BD and P A P C = P B P D. When is it true that ABCD is cyclic? Power of a Point is useful because it gives both a way to find a new cyclic quadrilateral, as well as use an old one. Obviously it is quite useful for things involving lengths. When are cyclic quadrilaterals useful? Some instances when they are likely to show up: Multiple points on a circle. Three points determine a circle; if you have more than this, then you have a bunch of equal angles for free. 3 If you don t know what this is, find out very soon. 18

8 Right angles. Pairs of right angles form cyclic quadrilaterals. (Why?) Not only that, but when you have right angles in a circle, this means you get diameters. Angles. Particularly equal ones. The problem is olympiad geometry. Maybe not so much the AIME, but if you re doing a decently hard olympiad problem, there s very often a cyclic quadrilateral somewhere Other Ways to Deal with Angles Here are two other useful facts. Fact. The sum of the angles in a triangle is 180. Fact. P lies on the opposite side of AB as C. P A is a tangent to the circumcircle of ABC if and only if P AB = ACB. P A B C Figure 1.3: Tangent Problems The usual warning about motivation here: you should not immediately go bananas marking angles every time you see a cyclic quadrilateral. Keep your eye on the prize. Despite how much I ve said this is a huge amount of beginning geometry, you will notice there are not that many problems here. In fact, what actually happened was that the problems section here got so large that I split most of it off into later sections, leaving only a few of the easier problems here. 1. Prove that cyclic trapezoids are isosceles, and vice-versa. 2. Let ABC be a triangle with altitudes AD, BE, CF meeting at the orthocenter H. (a) Among the seven points A, B, C, D, E, F and H, one can find six cyclic quadrilaterals. What are they? (b) Show that the altitudes of ABC are the angle bisectors of DEF. (c) (Almost nine-point circle) Point H is reflected over sides BC, CA and AB to points A 1, B 1 and C 1 in that order. Also, suppose HBCA 2, HCAB 2 and HABC 2 are parallelograms. Prove that ABCA 1 B 1 C 1 A 2 B 2 C 2 is a cyclic nonagon (probably self-intersecting). (d) (Nine-point circle) Prove that the circumcircle of DEF passes through the midpoints of the sides of ABC, as well as the midpoints of AH, BH and CH. 19

9 3. (Russia 1996) Points E and F are on side BC of convex quadrilateral ABCD (with E closer than F to B). It is known that BAE = CDF and EAF = F DE. Prove that F AC = EDB. 20

10 1.5 Fact 5 One particularly common configuration that you should know involves excircles. The A- excircle is the circle tangent to side BC and the extensions of sides AB and AC through B and C, respectively, and its center is the A-excenter. The B and C excircles and excenters are defined similarly. 1. Let ABC be a triangle with incenter I, and let I A, I B, and I C be the A, B and C-excenters. (a) Show that I B I C bisects the exterior angle of A. (b) Show that I is the orthocenter of triangle I A I B I C. 2. (Fact 5) Let ABC be a triangle. The incircle of ABC touches BC at a point D. The A-excircle of ABC touches AB, AC, and BC at points X B, X C and E, and has center J. Finally, the angle bisector of A hits the circumcenter of ABC at L. A I B D E C X C L X B J Figure 1.4: Fact 5 (a) Prove that A, I, L and J are collinear. (b) Compute AX B and AX C. (c) Show that BD = EC = s b and BE = DC = s c. (d) Show that L is the center of a circle passing through I, J, B and C. 21

11 1.6 TSTST 2012, Problem 7 To finish the geometry chapter, here s a documentation of my thought process against a fairly tough geometry problem as I solve it. This was not the solution I found during the test. Problem (TSTST 2012). Triangle ABC is inscribed in circle Ω. The interior angle bisector of angle A intersects side BC and Ω at D and L (other than A), respectively. Let M be the midpoint of side BC. The circumcircle of triangle ADM intersects sides AB and AC again at Q and P (other than A), respectively. Let N be the midpoint of segment P Q, and let H be the foot of the perpendicular from L to line ND. Prove that line ML is tangent to the circumcircle of triangle HM N. A Q N P B D M C H L Figure 1.5: Problem 7 Solution, with commentary. Well, first things first; that stupid tangent condition should be rewritten. Rewrite it as DNM = HML. Maybe right now you remember that BL = LC from an earlier problem; if not, the diagram should be a big hint. Anyways, it s not too hard to see that DML = 90. On the other hand, we are told DHL = 90. Bam, cyclic quad! (Told you these things show up everywhere.) So DMLH is cyclic, and now HDL = HML. So we just want to prove DNM = HDL. Wait a minute, that s just saying MN AD. And we know this has to be true for the problem to be true. But this means that we can erase a whole bunch of things: namely H, L, and a bunch of the lines related to them, leaving us which a much simpler problem. OK, so how do we prove MN AD? Hmm... 22

12 What s the hardest part of this problem? It s that darn circle passing through D and M, cutting the sides at P and Q, which at the moment we really know nothing about. What can we do?... And now here s the insight (aka the hard part of the problem). D and M are pretty good points in terms of lengths, so let s apply Power of a Point to try and get a grip on P and Q. We have BQ BA = BD BM CP CA = CD CM But BM = CM, and BD CD = BA CA. So this actually implies that BQ = CP! Some of you with better diagrams might have noticed this right off the bat, which is just as well. OK, so ( MN is now the average of these two equal lengths. Hmm... in vector language, MN = 1 2 BQ + CP ). But if BQ = CP, then BQ + CP is parallel to the angle bisector, namely AD, so MN must be as well. And suddenly the problem is solved! Now, I ve made the problem sound very easy. It s really not. I ended up throwing barycentric coordinates at this during the test after struggling with it, without success, for 90 minutes. But hopefully now you see that these solutions do come from somewhere, and what you should be thinking about while you re doing other problems. By the way, it is true that H lies on the Γ. In fact, if you let X be the second intersection of the two circles, it turns out that X lies on lines ML and NH. (Try and prove this!) This leads to a solution using an idea called spiral similarity, which some veterans would recognize instantly. (Remember the problem from the Angle Chasing section? Yes, same thing. It also appears in Yufei Zhao s Cyclic Quadrilaterals the Big Picture which I mentioned much earlier.) Here s a concise solution to the problem, and what should be submitted at an exam. Remember that solutions should be written forwards. Solution. First, we claim that MN AD. Notice that, by Power of a Point and the Angle Bisector Theorem, we have BQ = BD BM BA = CD CM CA = CP so BQ = CP. Now MN ( = 1 2 BQ + CP ), and since BQ = CP it follows that MN is parallel to the angle bisector of BAC, namely AD. Note that BL = LC DML = 90. Since DHL = 90, it follows that DMLH is cyclic. Now HMN = HDL = HML, and the conclusion follows. 23

13 1.7 The Champions Now that you have most of the tools, most problems are fair game. Good luck! 1. Let ω 1 and ω 2 be circles which intersect at points P and Q. AC is a chord of ω 1 and BD is a chord of ω 2. Suppose AC and BD intersect on P Q. Prove that ABCD is cyclic. 2. (JMO 2011) Points A, B, C, D, E lie on a circle ω and point P lies outside the circle. The given points are such that (i) lines P B and P D are tangent to ω, (ii) P, A, C are collinear, and (iii) DE AC. Prove that BE bisects AC. 3. (Mandelbrot 2012) Let ABC be a triangle inscribed in circle ω with BC = 17. The angle bisector of BAC intersects ω again at P. Given that sin ABP = 3 5, and that the radius of ω is 20, compute the area of quadrilateral ABP C. 4. (USAMO 2010) Let AXY ZB be a convex pentagon inscribed in a semicircle of diameter AB. Denote by P, Q, R, S the feet of the perpendiculars from Y onto lines AX, BX, AZ, BZ, respectively. Prove that the acute angle formed by lines P Q and RS is half the size of XOZ, where O is the midpoint of segment AB. 5. (IMO 2012) Given triangle ABC the point J is the centre of the excircle opposite the vertex A. This excircle is tangent to the side BC at M, and to the lines AB and AC at K and L, respectively. The lines LM and BJ meet at F, and the lines KM and CJ meet at G. Let S be the point of intersection of the lines AF and BC, and let T be the point of intersection of the lines AG and BC. Prove that M is the midpoint of ST. 6. (Brazil 2006) Let ABC be a triangle. The internal bisector of B meets AC in P. Let I be the incenter of ABC. Prove that if AP + AB = CB, then AP I is isosceles. 7. (Euler Line) In ABC, M is the midpoint of BC and AD is an altitude. Let H, G and O be the othrocenter, centroid, and circumcenter. (i) Prove that AM/GM = 3. (The AM-GM inequality will not help here.) (ii) Prove that AG/GM = 2. (iii) Let X and Y be the midpoints of AB and AC. By using similar triangles MXY and ABC, show that AH = 2OM. (iv) Prove that O, G, and H are collinear, and compute OG HG. (v) Prove that the center of the nine-point circle N is the midpoint of OH. 24

14 1.8 Quiz Problems 1. In quadrilateral ABCD, AB = 7, BC = 15, CD = 20, DA = 24, and BD = 25. Find CA. 2. In the figure, ABC is divided into six smaller triangles, four of whose areas are shown. Compute the area of ABC. C 84 A B 3. Let ABCDE be a convex pentagon such that BAC = CAD = DAE and ABC = ACD = ADE. Diagonals BD and CE meet at P. Prove that AP bisects CD. 4. Let ABCD be a cyclic quadrilateral. Denote by X and Y the orthocenters of ABD and ACD. Prove that BC = XY. 1. Let ABC be an acute triangle with circumcenter O, orthocenter H, and incenter I. Suppose that AH, BI, and CO concur and cos A cos B cos C = Compute 1000 cos B. 2. In triangle ABC, D is on AC and E is on AB such that: (i) Triangle ADE and quadrilateral BCDE have equal area and perimeter. (ii) Quadrilateral BCDE is cyclic. (iii) AD = 20 and CD = 12. The length of the altitude from A to BC can be expressed as k, for some positive integer k. Find k. 3. Let ABC be a triangle with incenter I. A point P in the interior of the triangle satisfies P BA + P CA = P BC + P CB. Show that AP AI, and that equality holds if and only if P = I. 25

15 1.8.1 Quiz 1 Solutions 1. In quadrilateral ABCD, AB = 7, BC = 15, CD = 20, DA = 24, and BD = 25. Find CA. Solution. Notice that ABCD is cyclic since A = C = 90. Theorem, AC = = 020. Then by Ptolemy s 2. (AIME 1985) In the figure, ABC is divided into six smaller triangles, four of whose areas are shown. Compute the area of ABC. C 84 A B Solution. We set the left blank area to y, and set the right blank area to x. Using the fact that triangles with similar altitudes, we see that if we have AB as the base, we get 124+y 65+x = y = 4x Using AC as the base, we get y 84 = 70+y 119+x. Solving, we obtain x = 70, y = 56. Hence, the total area is = (ISL 2006 G3, by Zuming Feng) Let ABCDE be a convex pentagon such that BAC = CAD = DAE and ABC = ACD = ADE. Diagonals BD and CE meet at P. Prove that AP bisects CD. Solution. Observe that ABC, ACD, and ADE are all similar. From this it is not hard to see that quadrilaterals ABCD and ACDE are similar. Denoting K = BD AC and L = CE AD, we find that AK KC = AL LC. The conclusion now follows via Ceva s Theorem. 4. (Evan Chen) Let ABCD be a cyclic quadrilateral. Denote by X and Y the orthocenters of ABD and ACD. Prove that BC = XY. Solution. (Aaron Lin) Reflect X and Y over side AD to points X and Y, which lie on the circumcircle of ABCD. Now notice that BCY X is a cyclic trapezoid; hence it is isosceles, so we have BC = X Y. Upon noticing XY = X Y we are done. 26

16 1.8.2 Quiz 2 Solutions 1. Let ABC be an acute triangle with circumcenter O, orthocenter H, and incenter I. Suppose that AH, BI, and CO concur and cos A cos B cos C = Compute 1000 cos B. Solution. Apply Trig Ceva to obtain sin (90 C) sin(b/2) sin (90 A) sin (90 B) sin(b/2) sin (90 B) = 1 This implies that cos 2 B = cos A cos C, so cos 3 B = cos A cos B cos C = cos B = (WOOT Mock AIME 2011, by David Stoner) In triangle ABC, D is on AC and E is on AB such that: (i) Triangle ADE and quadrilateral BCDE have equal area and perimeter. (ii) Quadrilateral BCDE is cyclic. (iii) AD = 20 and CD = 12. The length of the altitude from A to BC can be expressed as k, for some positive integer k. Find k. Solution. Since EBC = ADE we see that triangles ADE and ABC are similar. Furthermore, the ratio of their areas is 1 : 2, so the ratio of their side lengths is 1 : 2. It follows that AE = 1 2 AC = 16 2 and AB = 20 2, so that EB = 4 2. The condition on the perimeter now gives BC = AE + AD BE DC = From here we can compute the height from A to BC because we know all the side lengths of ABC. One approach is to compute [ABC] using Heron s Formula and divide by AB. Alternatively, let AL be the altitude, and write h = AL, x = BL and y = CL. Then we obtain the system: x 2 + h 2 = 800 y 2 + h 2 = 1024 x + y = Subtracting the first two equations gives y 2 x 2 = 224 y x = = , and from here we obtian x = 8, y = 12 2, so h 2 = (IMO 2006, by Hojoo Lee from South Korea) Let ABC be a triangle with incenter I. A point P in the interior of the triangle satisfies P BA + P CA = P BC + P CB. Show that AP AI, and that equality holds if and only if P = I. Solution. It is immediate that P BC + P CB = 1 2 (B + C), so BP C = 90 + A/2. Similarly, BIC = 90 + A/2, so BIP C is cyclic. Let AI meet the circumcircle of ABC at L. By Fact 5, P lies on a circle centered at L passing through B, I and C. But A, I and L are collinear, implying the conclusion. 27