Maths Olympiad Prep

Library / /3 of 69

Geometry Difficulty 4.8 AIME Prove it Mongolia

In a convex quadrilateral ABCDABCD, ∠B=∠C=120∘\angle B = \angle C = 120^\circ and AB2+BC2+CD2=AD2AB^2 + BC^2 + CD^2 = AD^2. Prove that there is a circle inscribed in ABCDABCD.

Solution

(AB+x)2+(CD+x)2−(AB+x)(CD+x)=AD2  ⟺  ⇒AB⋅x+CD⋅x−AB⋅CD=0. (AB + x)^2 + (CD + x)^2 - (AB + x)(CD + x) = AD^2 \iff \\ \Rightarrow AB \cdot x + CD \cdot x - AB \cdot CD = 0.
On the other hand, (AB+CD−x)2=AB2+CD2+x2−2(AB⋅x+CD⋅x−AB⋅CD)=AB2+CD2+x2=AD2(AB+CD-x)^2 = AB^2+CD^2+x^2-2(AB \cdot x+CD \cdot x-AB \cdot CD) = AB^2+CD^2+x^2 = AD^2 and we get AB+CD=AD+x=AD+BCAB+CD = AD+x = AD+BC. It means in the quadrilateral ABCDABCD can be inscribed a circle.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.