Olympiad Maths Prep

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Problem 696

AIME late
Geometry Difficulty 5.3 Prove it

14. Let PP be any point on the plane of rectangle ABCDA B C D. Prove that:
∣PA∣2+∣PC∣2=∣PB∣2+∣PD∣2. |P A|^{2}+|P C|^{2}=|P B|^{2}+|P D|^{2} .

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

14. As shown in the figure, ∣PA∣2+∣PC∣2=(x−a)2+(y−b)2+(x+a)2+(y+b)2=2(x2+y2+a2+b2),∣PB∣2|P A|^{2}+|P C|^{2}=(x-a)^{2}+(y-b)^{2}+(x+a)^{2}+(y+b)^{2}=2\left(x^{2}+y^{2}+a^{2}+b^{2}\right),|P B|^{2} +∣PD∣2=(x+a)2+(y−b)2+(x−a)2+(y+b)2=2(x2+y2+a2+b2)+|P D|^{2}=(x+a)^{2}+(y-b)^{2}+(x-a)^{2}+(y+b)^{2}=2\left(x^{2}+y^{2}+a^{2}+b^{2}\right). Therefore, ∣PA∣2+∣PC∣2=|P A|^{2}+|P C|^{2}= ∣PB∣2+∣PD∣2|P B|^{2}+|P D|^{2}.

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