IMO 1993 SL 14

Origin: ISR

Problem

The vertices D, E, F of an equilateral triangle lie on the sides BC, CA, AB respectively of a triangle ABC. If a, b, c are the respective lengths of these sides, and S the area of ABC, prove that DE \geq \sqrt 2S a2 + b2 + c2 + 4 \sqrt 3S .

Solution

Consider any point T inside the triangle ABC or on its boundary. Since 2S = 2(SAET F + SBF T D + SCDT E) \leqAT \cdot EF + BT \cdot FD + CT \cdot DE = (AT + BT + CT)DE, it suffices to find a point T such that (AT + BT + CT)2 \geqa2 + b2 + c2 + 4S \sqrt . We distinguish two cases: (i) If all angles of \triangleABC are less than 120◦, then the sum AT +BT +CT attains its minimum when T is the Torricelli point, i.e., such that \angleATB = \angleBTC = \angleCT A = 120◦. In this case, by the cosine theorem we get AT 2 + AT \cdot BT + BT 2 = c2, BT 2 + BT \cdot CT + CT 2 = a2, CT 2 + CT \cdot AT + AT 2 = b2, 3(AT \cdot BT + BT \cdot CT + CT \cdot AT ) = 4 \sqrt 3(SAT B + SBT C + SCT A) = 4 \sqrt 3S. Adding these four equalities, we obtain 2(AT + BT + CT )2 = a2 + b2 + c2 + 4 \sqrt 3S. (ii) Let \angleACB \geq120◦. We claim that T = C satisfies the requirements. Indeed, a2 + b2 + c2 + 4 \sqrt 3S = a2 + b2 + (a2 + b2 −2ab cos\angleC) + \sqrt 3ab sin \angleC = 2(a2 + b2) + 2ab( \sqrt 3 sin \angleC −cos \angleC) = 2(a2 + b2) + 4ab sin(\angleC −30◦) \leq2(a + b)2, which proves the desired inequality.