Maths Olympiad Prep

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, 2012

Geometry Difficulty 5.9 AIME, harder Prove it Belarus

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.

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.

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