HMMT 十一月 2010 · 冲刺赛 · 第 17 题

HMMT November 2010 — Guts Round — Problem 17

[ 9 ] A triangle with side lengths 5 , 7 , 8 is inscribed in a circle C . The diameters of C parallel to the sides of lengths 5 and 8 divide C into four sectors. What is the area of either of the two smaller ones?

解析

[ 9 ] A triangle with side lengths 5 , 7 , 8 is inscribed in a circle C . The diameters of C parallel to the sides of lengths 5 and 8 divide C into four sectors. What is the area of either of the two smaller ones? 49 Answer: π Let 4 P QR have sides p = 7, q = 5, r = 8. Of the four sectors determined by 18 the diameters of C that are parallel to P Q and P R , two have angles equal to P and the other two have angles equal to π − P . We first find P using the law of cosines: 49 = 25 + 64 − 2(5)(8) cos P 1 π π implies cos P = implies P = . Thus the two smaller sectors will have angle . Next we find the 2 3 3 pqr circumradius of 4 P QR using the formula R = , where [ P QR ] is the area of 4 P QR . By Heron’s 4[ P QR ] √ √ 5 · 7 · 8 7 √ √ Formula we have [ P QR ] = 10(5)(3)(2) = 10 3; thus R = = . The area of a smaller sector 4(10 3) 3 ( ) 2 ( ) π/ 3 2 π 7 49 √ is thus πR = = π 2 π 6 18 3