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Geometry Difficulty 7.7 National Olympiad, round 2 Prove it Hong Kong

Let AA, BB and CC be real numbers such that
(i) sinAcosB+cosAsinB=sinAcosA+sinBcosB\sin A \cos B + |\cos A \sin B| = \sin A |\cos A| + |\sin B \cos B|,
(ii) tanC\tan C and cotC\cot C are defined.
Find the minimum value of (tanCsinA)2+(cotCcosB)2(\tan C - \sin A)^2 + (\cot C - \cos B)^2.

Solution

The minimum value is 3223 - 2\sqrt{2}.
Condition (i) can be rewritten as
(sinAsinB)(cosBcosA)=0. (\sin A - |\sin B|)(\cos B - |\cos A|) = 0.
If sinA=sinB\sin A = |\sin B|, then we have sin2A=sin2B=1cos2B\sin^2 A = \sin^2 B = 1 - \cos^2 B.
If cosB=cosA\cos B = |\cos A|, then we have cos2B=cos2A=1sin2A\cos^2 B = \cos^2 A = 1 - \sin^2 A.
Therefore, in any case, (sinA,cosB)(\sin A, \cos B) is a point on the unit circle on the coordinate plane. Since the steps are reversible, it can be any point on the unit circle.
Figure 1
Next, (tanC,cotC)(\tan C, \cot C) can be any point on the hyperbola xy=1xy = 1. Therefore, we are asked to find the minimum distance between a point on the unit circle and a point on the hyperbola. In view of the geometry, since the line y=xy = x is an axis of symmetry of both geometric objects, the minimum distance is attained

between points on the line y=xy = x. By symmetry, it suffices to consider x>0x > 0. Clearly, (12,12)\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) is a solution to x2+y2=1x^2 + y^2 = 1 and y=xy = x, while (1,1)(1, 1) is a solution to xy=1xy = 1 and y=xy = x. Therefore, the minimum distance is
(112)2+(112)2=322, \left(1 - \frac{1}{\sqrt{2}}\right)^2 + \left(1 - \frac{1}{\sqrt{2}}\right)^2 = 3 - 2\sqrt{2},
and we are done.

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