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Problem 1862

National Olympiad, first round

Geometry Difficulty 6.3 Prove it Iberoamerican Mathematical Olympiad · Ibero-American Mathematical Olympiad

$\mathrm{ABCD}$ is a square. $\mathrm{P}, \mathrm{Q}$ are points on the sides $\mathrm{BC}, \mathrm{CD}$ respectively, distinct from the endpoints such that $\mathrm{BP}=\mathrm{CQ}$ . $\mathrm{X}, \mathrm{Y}$ are points on $\mathrm{AP}, \mathrm{AQ}$ respectively. Show that there is a triangle with side lengths $BX, XY, YD$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

We have $DY < BY \leq BX + XY$ (this is almost obvious, but to prove formally use the cosine formula for $BAY$ and $DAY$ and notice that $\angle BAY > \angle DAY$ ). Similarly, $BX < DX \leq DY + YX$ . So it remains to show that $XY < BX + DY$ .

Take $Q'$ on the extension of $BC$ so that $BQ' = DQ$ , as shown in the diagram. Take $Y'$ on $AQ'$ so that $AY' = AY$ . Then $XY' \leq BX + BY' = BX + DY$ . Now we claim that $\angle PAQ' > \angle PAQ$ , so it follows by the same observation as above that $XY' > XY$ . But the claim is almost obvious. Note that $PQ' = AB$ .

So take $P'$ on $AD$ with $\angle P'PQ' = 90^\circ$ . Then $A$ lies inside the circle $P'PQ'$ , so extend $PA$ to meet it again at $A'$ . Then $\angle PA'Q' = \angle PP'Q' = 45^\circ$ , so $\angle PAQ' = \angle PA'Q' + \angle AQ'Q' > 45^\circ$ . But $\angle PAQ' + \angle PAQ = 90^\circ$ , so $\angle PAQ' > \angle PAQ$ as claimed.

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