Similar Quadrilaterals

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1 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 1 Similar Quadrilaterals uthors Guangqi ui, kshaj Kadaveru, Joshua Lee, Sagar Maheshwari Special thanks to osmin Pohoata and the MSP ornell 2014 Geometric Proofs lass dditional thanks to Justin Stevens and avid ltizio for the L TEX Template

2 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 2 ontents 1 Introduction 3 2 Interesting Property 4 3 Example Problems 5 4 Practice Problems 11

3 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 3 1 Introduction Similar quadrilaterals are a very useful but relatively unknown tool used to solve olympiad geometry problems. It usually goes unnoticed due to the confinement of geometric education to the geometry of the triangle and other conventional methods of problem solving. lso, it is only in very special cases where pairs of similar quadrilaterals exist, and proofs using these qualities usually shorten what would have otherwise been an unnecessarily long proof. The most common method of finding such quadrilaterals involves finding one pair of adjacent sides with identical ratios, and three pairs of congruent angles. We will call this SS Similarity. Example 1.1. (SS Similarity) Two quadrilaterals and satisfy =, =, =, and =. Show that and are similar. Solution. Notice and are similar from SS similarity. Therefore = = =. Similarly = so. Now notice we have = = = = so our proof is complete.

4 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 4 2 Interesting Property Example 2.1. Let be a cyclic quadrilateral. We define G, H, N to denote the centroid, orthocenter, and nine-point center of. We define G, etc. similarly. Show that, H H H H, G G G G, N N N N are all similar quadrilaterals with a similarity ratio of 6 : 6 : 2 : 3, respectively. Figure 2.1: iagram made by Evan hen Solution. From a well-known lemma, H = 2R cos. We know 2R cos = 2R cos = H lso H H, so H H is a parallelogram, so H H, = H H. Similarly, H H, H H, H H, and so is similar to H H H H, and furthermore congruent. It is well known that the nine-point center N lies on the Euler line, and is the midpoint between the circumcenter O and the orthocenter H a. Therefore N N N N is simply a dilation of H H H H by a factor of two, meaning it is similar to H H H H and therefore similar to with a ratio of 1 2. It is well-known that OG : OH = 1 : 3, so G G G G is similar to H H H H with a ratio of 1 3. Since H H H H is congruent to, we get our result.

5 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 5 3 Example Problems Example 3.1. (Salmon s Lemma) Let O be the center of an arbitrary circle and let P, Q be arbitrary points. If a is the distance from P to the polar of Q, and b is the distance from Q to the polar of P, then show that OP OQ = a b. Y b Q X Q a P P O Solution. raw the tangents from P, Q to circle O to intersect at,,, respectively. Notice is the polar of P, and is the polar of Q. rop perpendiculars from P, Q to, at X, Y respectively. Let P O intersect at P, and QO intersect at Q. Notice that because is the polar of P, we have P P = P O = 90. From similar triangles P O P O, we have OP OP = O 2 = R 2, and similarly from right triangle OQ we have OQ OQ = O 2 = R 2, so OP OP = OQ OQ. Notice quadrilaterals XP OQ and Y QOP are similar because they have: two right angles each at X, Q and Y, P ; share an angle at O; and furthermore satisfy OQ = OP OP. From this we get OP OQ OQ = P X QY = a as desired. b

6 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 6 Example 3.2. In a triangle, construct altitudes, E, F and let H be the orthocenter. Let O 1, O 2, O 3 be the incenters of triangles EHF, F H, HE, respectively. Prove that the lines O 1, O 2, O 3 are concurrent. E F O 1 H O 1 H 1 Solution. enote H 1 as the point outside triangle where H 1 = F EH and H 1 = EF H. enote the incenter of the triangle H 1 as O 1. efine O 2 and O 3 similarly. Then, by Jacobi s Theorem, we see that O 1, O 2, O 3 are concurrent at a point P. Notice that F because F = E = 90, F E is cyclic so from Power of a Point, =. lso, E F E =, EF = so F H = H 1 and EH = H 1. So from SS similarity, we know quadrilaterals EHF and H 1 are similar. Therefore O 1 = O 1. Similarly, we get O 2 = O 2 and O 3 = O 3, so O 1, O 2, O 3 are concurrent at the isogonal conjugate of P.

7 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 7 Next, we will prove an intriguing lemma that will be key to a more difficult problem. oth the lemma and the following problem will use properties of similar quadrilaterals in their proofs. Example 3.3. Let 1 2 n and 1 2 n be directly similar n-gons, and k be a real number. Show that 1 2 n is similar to both 1 2 n and 1 2 n, where for 1 i n, i is on i i with ratio i i : i i = k. Solution. onsider the center of spiral similarity X of 1 2 n and 1 2 n. Therefore all triangles of the form i X i for 1 i n are similar. We know all degenerate quadrilaterals of the form X i i i have i i : i i, X i i, i i i, and X i i constant, because they are equal to k, X i i, 180, and X i i, respectively, which are known to be constant. Therefore, from SS similarity, we have for all 1 i n, the quadrilaterals X i i i are similar. It is now clear that the ratio X i X i and the angle i X i are constant over i, meaning there exists a spiral similarity mapping 1 2 n to 1 2 n implying the result. Surprisingly, this property also holds for k similar n-gons, where i is defined as the centroid of polygon i i, obtained by averaging the cartesian coordinates of the vertices. Now we proceed to the original problem. Example 3.4. (US TST 2000) Let be a cyclic quadrilateral and let E and F be the feet of perpendiculars from the intersection of diagonals and to and, respectively. Prove that EF is perpendicular to the line through the midpoints of and. X E M 1 P F Y M 2 Solution. enote P as the intersection of the diagonals and M 1 and M 2 as the midpoints of and, respectively. Let X and Y be the reflections of P across and, respectively. Since = and =, triangles P and P are similar, which implies

8 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 8 quadrilaterals P X and Y P are similar. Notice quadrilateral M 1 EM 2 F is formed by the midpoints of, P Y,, and P X, so from the preceding lemma, this quadrilateral is similar to both P X and Y P. ecause P X and Y P are kites, M 1 F M 2 E is also a kite which implies M 1 M 2 EF as desired. Example 3.5. (ISL 2009) Given a cyclic quadrilateral, let the diagonals and meet at E and the lines and meet at F. The midpoints of and are G and H, respectively. Show that EF is tangent at E to the circle through the points E, G and H. F G E H G H Solution. Let G be the reflection of point F across point G, and let H be the reflection of point F across point H. Note that we have parellelograms F G and F H, which are similar because triangles F and F are similar. From this and from E E we obtain that quadrilaterals F E and G E are similar, and the quadrilaterals F E and H E are similar. These pairs of similar quadrilaterals implies that the position of point E in F G corresponds to the same position as point E in F H. Therfore, we have that G EF = H EF which implies that E, G, H are collinear. ecause G H is a homothetic transformation of scale 2 of GH in respect to the point F, we see that GH and G H are parallel. This means that GHE = H EH, and since we know that F EG = H EH, we have F EG = GHE implying F E is a tangent to the circumcircle of GEH at E.

9 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 9 It is important to realize that similar quadrilaterals often do not present themselves. s you saw in the past few problems, you should try to find clever reflections and transformations which generate useful similar quadrilaterals. Occasionally, drawing auxiliary lines can also aid in creating similar quadrilaterals. Example 3.6. (IMO 2013) Let be an acute triangle with orthocenter H, and let W be a point on the side, lying strictly between and. The points M and N are the feet of altitudes from and, respectively. enote by ω 1 the circumcircle of W N, and let X be the point such that W X is a diameter of ω 1. nalogously, denote by ω 2 the circumcircle of W M, and let Y be the point such that W Y is a diameter of ω 2. Prove that X, Y, and H are collinear. ω 2 Y N M ω 1 H X W Solution. Note that N = HN = 90, and = 90 M = NH, so NH N. lso, we know NX = 180 NW = NW, and because W X is a diameter, XN = 90 NW = W N. Therefore, XN W N. From both similar triangles we get N = NH, NX = NW, NW = N+ NW = HN+ NX = XNH, and N = NH. Therefore, from SS, quadrilaterals XNH and W N are similar. H From this we obtain NHX = NW = W. We analogously obtain quadrilaterals Y MH and W M are similar, meaning W = Y H, and we conclude that NHX = Y H. Therefore points X, H, Y are collinear. Similar quadrilaterals can provide an amazing shortcut to a difficult problem like this.

10 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 10 Example 3.7. (IMO 2004) In a convex quadrilateral, the diagonal bisects neither the nor the. The point P lies inside and satisfies P =, Prove that is a cyclic quadrilateral if P = P. P =. Y P Solution. ssume / (). Let X = P () and Y = P (). We shall prove that quadrilaterals P and P Y X are similar, and furthermore congruent. y angle-chasing: P = = Y, P = = X, and P = Y P X. lso, by Power of a Point P. So from SS similarity we can conclude that P P Y X. ut the = P Y P P X similarity ratio between the two is P which equals one, so the two quadrilaterals are congruent. P In fact they are reflections of each other over the angle bisector of P X. Since Y X is cyclic its reflection Y X is cyclic as well. ut these two circles are the same implying that is cyclic. X

11 Similar Quadrilaterals ui, Kadaveru, Lee, Maheshwari Page 11 4 Practice Problems Now that you have seen various solutions using similar quadrilaterals, try it yourself! See if you can apply similar quadrilaterals to the following problems: 1. (India 2014) In a acute-angled triangle, a point lies on the segment. Let O 1, O 2 denote the circumcentres of triangles and respectively. Prove that the line joining the circumcentre of triangle and the orthocentre of triangle O 1 O 2 is parallel to. 2. (Korea 2010) Let be a cyclic convex quadrilateral. Let E be the intersection of lines,. P is the intersection of line passing and perpendicular to, and line passing and perpendicular to. Q is the intersection of line passing and perpendicular to, and line passing and perpendicular to. Prove that three points E, P, Q are collinear. 3. (US 2006) Let be a triangle. Triangles P and Q are constructed outside of triangle such that P = and Q = and P = Q. Segments Q and P meet at R. Let O be the circumcenter of triangle R. Prove that O P Q. 4. (ISL 2003) Let be an isosceles triangle with =, whose incentre is I. Let P be a point on the circumcircle of the triangle I lying inside the triangle. The lines through P parallel to and meet at and E, respectively. The line through P parallel to meets and at F and G, respectively. Prove that the lines F and EG intersect on the circumcircle of the triangle. 5. (Greece 2001) triangle is inscribed in a circle of radius R. Let and E be the bisectors of the angles and respectively and let the line E meet the arc not containing at point K. Let 1, 1, 1 be the feet of perpendiculars from K to,,, and x, y be the distances from and E to, respectively. (a) Express the lengths of K 1, K 1, K 1 in terms of x, y and the ratio l = K/E. (b) Prove that 1 K = 1 K + 1 K.

Example G1: Triangles with circumcenter on a median. Prove that if the circumcenter of a triangle lies on a median, the triangle either is isosceles

1 Example G1: Triangles with circumcenter on a median. Prove that if the circumcenter of a triangle lies on a median, the triangle either is isosceles or contains a right angle. D D 2 Solution to Example