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Problem 964

AIME late
Geometry Difficulty 5.9 Prove it Belorusija 2012 · Belarus · 2012

Point MM is marked inside the convex quadrilateral ABCDABCD so that the ratio of the areas of the triangles AMCAMC and BMDBMD is equal to the ratio of the tangents of the angles AMCAMC and BMDBMD, i.e. S(AMC):S(BMD)=tg⁡∠AMC:tg⁡∠BMDS(AMC) : S(BMD) = \tg \angle AMC : \tg \angle BMD.
Prove that AM2+MC2+BD2=AC2+BM2+MD2AM^2 + MC^2 + BD^2 = AC^2 + BM^2 + MD^2 if MM does not belong to any of the diagonals of the quadrilateral.

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

Since
S(AMC)=0.5AM⋅MCsin⁡∠AMC,S(BMD)=0.5BM⋅MDsin⁡∠BMD, S(AMC) = 0.5 AM \cdot MC \sin \angle AMC, \quad S(BMD) = 0.5 BM \cdot MD \sin \angle BMD,
we have
AM⋅MC=2S(AMC)/sin⁡∠AMC,BM⋅MD=2S(BMD)/sin⁡∠BMD.(1) AM \cdot MC = 2S(AMC)/\sin \angle AMC, \quad BM \cdot MD = 2S(BMD)/\sin \angle BMD. \quad (1)

Figure 1

By the cosine law,
AC2=AM2+MC2−2AM⋅MCcos⁡∠AMC, AC^2 = AM^2 + MC^2 - 2AM \cdot MC \cos \angle AMC,
BD2=BM2+MD2−2BM⋅MDcos⁡∠BMD. BD^2 = BM^2 + MD^2 - 2BM \cdot MD \cos \angle BMD.

From (1) it follows
AC2=AM2+MC2−4S(AMC)cos⁡∠AMC/sin⁡∠AMC=AM2+MC2−4S(AMC)/tg⁡∠AMC,(2) AC^2 = AM^2 + MC^2 - 4S(AMC) \cos \angle AMC / \sin \angle AMC = \\ AM^2 + MC^2 - 4S(AMC) / \operatorname{tg} \angle AMC, \qquad (2)
BD2=BM2+MD2−4S(BMD)cos⁡∠BMD/sin⁡∠BMD=BM2+MD2−4S(BMD)/tg⁡∠BMD.(3) BD^2 = BM^2 + MD^2 - 4S(BMD) \cos \angle BMD / \sin \angle BMD = \\ BM^2 + MD^2 - 4S(BMD) / \operatorname{tg} \angle BMD. \qquad (3)
By condition,
S(AMC)/tg⁡∠AMC=S(BMD)/tg⁡∠BMD, S(AMC) / \operatorname{tg} \angle AMC = S(BMD) / \operatorname{tg} \angle BMD,
so (2) and (3) gives the required equality.

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