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Extra resources for 44th International Mathematical Olympiad: Short-listed problems and solutions
Then ∠AP B = ∠AP Q + ∠BP Q = ∠P DA + ∠P CB. Define θ1 , . . , θ8 as in Figure 1. Then θ2 + θ3 + ∠AP B = θ2 + θ3 + θ5 + θ8 = 180◦ . (1) Similarly, ∠BP C = ∠P AB + ∠P DC and θ4 + θ5 + θ2 + θ7 = 180◦ . (2) Multiply the side-lengths of the triangles P AB, P BC, P CD, P AD by P C ·P D, P D·P A, P A · P B, P B · P C, respectively, to get the new quadrilateral A B C D as in Figure 2. 37 DA · P B · P C D A θ1 θ6 θ8 PB · PC · PD θ3 PA · PB · PC P CD · P A · P B AB · P C · P D PC · PD · PA PD · PA · PB θ2 θ7 θ4 θ5 BC · P D · P A C B Figure 2 (1) and (2) show that A D B C and A B C D .
In fact, the common value of AP 2 + P D2 , BP 2 + P E 2 , CP 2 + P F 2 is equal to 8R2 − s2 , where R is the circumradius of the triangle ABC and s = (BC + CA + AB)/2. We can prove this as follows: Observe that the circumradius of the triangle IA IB IC is equal to 2R since its orthic triangle is ABC. It follows that P D = P IA − DIA = 2R − rA , where rA is the radius of the excircle of the triangle ABC opposite to A. Putting rB and rC in a similar manner, we have P E = 2R − rB and P F = 2R − rC .
Set the coordinate system as in the following figure. 40 y D Γ4 Γ1 θ C x A B θ Γ2 Γ3 We may assume that Γ1 : x2 + y 2 + 2ax sin θ − 2ay cos θ = 0, Γ3 : x2 + y 2 − 2cx sin θ + 2cy cos θ = 0, Γ2 : x2 + y 2 + 2bx sin θ + 2by cos θ = 0, Γ4 : x2 + y 2 − 2dx sin θ − 2dy cos θ = 0. Simple computation shows that A B C D 4ab(a + b) sin θ cos2 θ 4ab(a − b) sin2 θ cos θ ,− 2 , − 2 a + b2 + 2ab cos 2θ a + b2 + 2ab cos 2θ 4bc(b − c) sin θ cos2 θ 4bc(b + c) sin2 θ cos θ ,− 2 , b2 + c2 − 2bc cos 2θ b + c2 − 2bc cos 2θ 4cd(c + d) sin θ cos2 θ 4cd(c − d) sin2 θ cos θ , , c2 + d2 + 2cd cos 2θ c2 + d2 + 2cd cos 2θ 4da(d − a) sin θ cos2 θ 4da(d + a) sin2 θ cos θ , .