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Nonetheless, the argument given above remains valid. Students can show that O lies outside the triangle if and only if one of its angles is greater than 120◦ . One argument might come from noting that BB and CC meet at a 120◦ angle no matter what the angles of the original triangle are. If, say, the triangle’s interior angle at A is greater than 120◦ , then point A must lie inside the circle through B, O, C, so that O lies outside the triangle. 46 Problem 364. More generally, find a point such that the sum of its distances to the three vertices of a triangle, multiplied by given positive numbers , m, n, is minimal.

By direct K 0 L L0 J computation we have (in magnitude and sign) JJ00K L · K0 J · L0 K = −1, so that (198) lines JJ0 , KK0 , LL0 are concurrent at some point O0 . Also by direct computation, we have J0 K L0 J JK0 = · . K0 L J0 L L0 K Now we apply Menelaus’ theorem (192) to triangle KJJ0 with transversal LL0 , to L0 K 0 get OO00JJ0 · LJ LK · L0 J = 1, which we can write as (11) JO0 JL0 LK JL0 LJ0 + J0 K JL0 J0 K = · = · = · 1+ O0 J0 L0 K LJ0 L0 K LJ0 L0 K LJ0 JL0 J0 K JL0 J0 K L0 J JL0 + · = + · .

Indeed, equation (6) shows that, for example, if we fix K , L , a smaller length for J J1 results in a larger value for s. 3◦ . In this case, s has neither a maximum nor a minimum, for any triangle J K L . Indeed, suppose, for example, m + n > 0, n + < 0. Then an examination of equation (6) shows that if we fix K , L , and move J away from the foot of the perpendicular from J1 (along line BC), then s keeps increasing. But if we fix J , L , and move K1 (along line AC), then s keeps decreasing. 4◦ .

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