Track / Stage 6 / 213 of 400 #1213 of 2000

Problem 1213

National olympiad, first round

Geometry Difficulty 6.3 Prove it

(IMO 2022 P4) Let $A B C D E$ be a convex pentagon such that $B C=D E$ . Suppose there exists a point $T$ inside $A B C D E$ such that $T B=T D, T C=T E$ and $\widehat{A B T}=\widehat{A E T}$ . Let $P$ and $Q$ be the points of intersection of the lines $(C D)$ and $(C T)$ with the line $(A B)$ ; assume that the points $P, B, A$ and $Q$ are collinear in this order. Similarly, let $R$ and $S$ be the points of intersection of the lines $(C D)$ and $(D T)$ with the line $(A E)$ , and assume that the points $R, E, A$ and $S$ are also collinear in this order. Prove that the points $P, S, Q$ and $R$ are concyclic.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

To draw the figure, we start with the segment $[A T]$ . We take a point $B$ and denote $B^{\prime}$ as its symmetric point with respect to the segment $[A T]$ . We can then choose $E$ on the circumcircle of triangle $A B^{\prime} T$ . Next, we choose a point $C$ on the circle centered at $T$ with radius $T E$ . The point $D$ will then be the intersection of the circle centered at $T$ with radius $T B$ and the circle centered at $E$ with radius $B C$ .

Now let's proceed to the solution of the exercise. The triangles $ETD$ and $CTB$ are isometric according to the length conditions. We deduce that

$\widehat{Q T E}=180^{\circ}-\widehat{E T D}-\widehat{D T C}=180^{\circ}-\widehat{B T C}-\widehat{D T C}=\widehat{S T B}$

If we denote $X$ and $Y$ as the points of intersection of $(Q T)$ with $(A E)$ and of $(S T)$ with $(A B)$ , respectively, then we have

$\widehat{E X T}=180^{\circ}-\widehat{X E T}-\widehat{X T E}=180^{\circ}-\widehat{T Y B}-\widehat{Y T B}=\widehat{T Y B}$

so that $\widehat{Q X S}=\widehat{Q Y S}$ and the points $Q, X, Y$ , and $S$ are concyclic.

On the other hand, since triangles $E X T$ and $B Y T$ share the same angles pair by pair, they are similar and

$\frac{X T}{Y T}=\frac{E T}{B T}=\frac{T C}{T D}$

so that by Thales' theorem, the lines $(X Y)$ and $(C D)$ are parallel. We can then conclude

$\widehat{P Q S}=\widehat{Y Q S}=\widehat{Y X S}=\widehat{Y X A}=\widehat{P R A}=\widehat{P R S}$

therefore the points $P, R, Q$ , and $S$ are concyclic.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.