ABCD is a square. P, Q are points on the sides BC, CD respectively, distinct from the endpoints such that BP = CQ. X, Y are points on AP, AQ respectively. Show that there is a triangle with side lengths BX, XY, YD.
Solution
We have DY < BY ≤ BX + XY (this is almost obvious, but to prove formally use the cosine formula for BAY and DAY and notice that ∠BAY > ∠DAY). Similarly, BX < DX ≤ 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' ≤ BX + BY' = BX + DY. Now we claim that ∠PAQ' > ∠ 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 ∠P'PQ' = 90o. Then A lies inside the circle P'PQ', so extend PA to meet it again at A'. Then ∠PA'Q' = ∠PP'Q' = 45o, so ∠PAQ' = ∠PA'Q' + ∠AQ'Q' > 45o. But ∠PAQ' + ∠PAQ = 90o, so ∠ PAQ' > ∠ PAQ as claimed.
© John Scholes
jscholes@kalva.demon.co.uk
30 Dec 03
Last corrected 30 Dec 03