Let $ABC$ be a triangle with $\angle B=90$. The incircle of $ABC$ touches the side $BC$ at $D$. The incenters of triangles $ABD$ and $ADC$ are $X$ and $Z$ , respectively. The lines $XZ$ and $AD$ are intersecting at the point $K$. $XZ$ and circumcircle of $ABC$ are intersecting at $U$ and $V$. Let $M$ be the midpoint of line segment $[UV]$ . $AD$ intersects the circumcircle of $ABC$ at $Y$ other than $A$. Prove that $|CY|=2|MK|$ .
Problem
Source: Turkey NMO 2007 Problem 5
Tags: geometry, incenter, circumcircle, inradius, rectangle, geometry unsolved
sunken rock
28.09.2011 00:18
Hint: key point is to show that the 2 incircles are tangent to each other. Best regards, sunken rock
Vo Duc Dien
07.10.2011 06:29
Easy said than done! I'm writing up my solution now.
Vo Duc Dien
07.10.2011 08:29
Let the incircle of triangle ABC touch the side AB at I, H and F be the feet of Z onto AD and DC, respectively, N and E the feet of X onto AB and BD, respectively, and L be the midpoint of AY. We need to show that the two incircles of triangles ABD and ACD are tangent to each other at H or K. To do this we need to prove that AH = AN. It’s easily seen that AC + AD + CD = 2(AH + DH + CF), or AH = ½ (AC + AD + CD) – (DH + CF) = ½ (AC + AD – CD). Similarly, AN = ½ (AB + AD – BD), and AH = AN when AB + CD = AC + BD, but AC = AI + CD, and the previous equation becomes AB = AI + BD (i) Since BD = BI is the inradius of triangle ABC, and the equation (i) is true. Therefore, the two incircles of triangles ABD and ACD are tangent to each other at H or K, or AK⊥UV. Since M is the midpoint of UV and O is the circumcenter, OM ⊥UV. Combining with AK⊥UV, we have LK || OM. It’s also because ∠B = 90°, AC is the diameter of the circumcircle and ∠AYC = 90°. Since O and L are the midpoints of AC and AY, respectively, OL || CY and OL = ½ CY, and thus we also have OL⊥AY, and OMKL is now a rectangle which implies that OL = MK. Therefore, we finally have |CY| = 2|MK|.