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2001 USAMO Problems/Problem 2

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Problem

Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$ , respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$ , respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$ , and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$ . Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$ . Prove that $AQ = D_2P$ .

Solution

It is well known that the excircle opposite $A$ is tangent to $\overline{BC}$ at the point $D_2$ . (Proof: let the points of tangency of the excircle with the lines $BC, AB, AC$ be $D_3, F,G$ respectively. Then $AB+BD_3=AB + BF=AF = AG = AC + AG=AC + CD_3$ . It follows that $2CD_3 = AB + BC - AC$ , and $CD_3 = s-b = BD_1 = CD_2$ , so $D_3 \equiv D_2$ .)

Now consider the homothety that carries the incircle of $\triangle ABC$ to its excircle. The homothety also carries $Q$ to $D_2$ (since $A,Q,D_2$ are collinear), and carries the tangency points $E_1$ to $G$ . It follows that $\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}$ .

pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6);pair B=(0,0), C=(10,0), A=7*expi(1),O...

By Menelaus' Theorem on $\triangle ACD_2$ with segment $\overline{BE_2}$ , it follows that $\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(c - (s-a)) \cdot a...$ . It easily follows that $AQ = D_2P$ . $\blacksquare$

See also

2001 USAMO (Problems • Resources: AoPS \| ML)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2001_USAMO_Problems/Problem_2"

Category: Olympiad Geometry Problems

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