Menelaus and Length Bashing

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Menelaus and Length Bashing

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Source: Korean National Junior Olympiad Number 7

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#1 h Nov 2, 2014, 9:28 AM • 2 Y

Y by Adventure10, Mango247

In a parallelogram $\Box ABCD$ $(AB < BC)$
The incircle of $\triangle ABC$ meets $\overline {BC}$ and $\overline {CA}$ at $P, Q$ .
The incircle of $\triangle ACD$ and $\overline {CD}$ meets at $R$ .
Let $S$ = $PQ$ $\cap$ $AD$
$U$ = $AR$ $\cap$ $CS$
$T$ , a point on $\overline {BC}$ such that $\overline {AB} = \overline {BT}$

Prove that $AT, BU, PQ$ are concurrent

This post has been edited 1 time. Last edited by rkm0959, Jun 7, 2015, 2:51 AM

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#2 h Nov 2, 2014, 2:05 PM • 1 Y

Y by Adventure10

My solution:

Let $X=AT \cap PQ$ .
Let $I$ be the incenter of $\triangle ABC$ .
Let $L_S$ be a line passing through $S$ and parallel to $AB$ .
Let $L_R$ be a line passing through $R$ and parallel to $BC$ .
Let $S'=L_S \cap BC, R'=L_R \cap AB, K=L_S \cap L_R$ .
Let $Y_{\infty}$ be the infinity point with direction $CD$ .
Let $Z_{\infty}$ be the infinity point with direction $AD$ .

Since it's well known that $B, I, X$ are collinear ,
so it is suffices to prove $B, I, U$ are collinear .
Since $KR'=AS=AQ=CR=KS'$ ,
so we get $KS'BR'$ is a rhombus and $B, I, K$ are collinear. ... $(1)$
From Pappus theorem (for $\{ S,A, Z_{\infty} \}$ and $\{ R, C, Y_{\infty} \}$ ) we get $B, K, U$ are collinear. ... $(2)$
From $(1)$ and $(2)$ we get $B, I, U$ are collinear .

Q.E.D

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#3 h Oct 27, 2016, 4:26 PM • 2 Y

Y by Elcin_bayramli, Adventure10

Any idea with application Menelaus theorem.?

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