PUMaC 2012 · 几何（A 组） · 第 7 题
PUMaC 2012 — Geometry (Division A) — Problem 7
题目详情
- [ 7 ] An octahedron (a solid with 8 triangular faces) has a volume of 1040. Two of the spa- tial diagonals intersect, and their plane of intersection contains four edges that form a cyclic quadrilateral. The third spatial diagonal is perpendicularly bisected by this plane and inter- sects the plane at the circumcenter of the cyclic quadrilateral. Given that the side lengths of the cyclic quadrilateral are 7 , 15 , 24 , 20, in counterclockwise order, the sum of the edge lengths of the entire octahedron can be written in simplest form as a/b . Find a + b .
解析
- [ 7 ] An octahedron (a solid with 8 triangular faces) has a volume of 1040. Two of the spa- tial diagonals intersect, and their plane of intersection contains four edges that form a cyclic quadrilateral. The third spatial diagonal is perpendicularly bisected by this plane and inter- sects the plane at the circumcenter of the cyclic quadrilateral. Given that the side lengths of the cyclic quadrilateral are 7 , 15 , 24 , 20, in counterclockwise order, the sum of the side lengths of the entire octahedron can be written in simplest form as a/b . Find a + b . 2 2 2 2 Solution: Note that 7 + 24 = 15 + 20 , so the cyclic quadrilateral has perpendicular diag- onals. For a quadrilateral with perpendicular diagonals, the area is K = pq/ 2 where p and q are the lengths of the diagonals. Thus we can compute the area using Ptolemy’s theorem: K = pq/ 2 = ( ac + bd ) / 2 where a, b, c, d are consecutive lengths of the sides. Plugging in, we have the area K = 234. The volume of the octahedron is the sum of the volumes of two pyramids, each with the quadrilateral as the base and half of the third diagonal as the height. So volume of the octahedron is V = KL/ 3 where L is the length of the third diagonal. Solving for L , the length of the third spatial diagonal is 3 ∗ 1040 / 234 = 40 / 3. 4 Label the quadrilateral as ABCD (such that a = AB , b = BC , c = CD , and d = DA ), and let O be the center of the circumcircle of this quadrilateral. To get the circumradius, we consider ◦ the fact that the arcs of opposite sides add up to 180 because the diagonals are perpendicular. This implies that ∠ AOB is supplementary to ∠ COD , so cos ∠ AOB = − cos ∠ COD . Using the 2 2 2 law of cosines on 4 AOB and 4 COD and adding, we have 4 R = a + c . So the circumradius √ 2 2 is R = ( a + c ) / 4 = 25 / 2. Using the pythagorean theorem, we can find each missing side length by taking R to be one leg, half of L to be the other leg, and the missing side length to be the hypoteneuse. Thus each missing side length has a length of 85 / 6, and the whole octahedron has 8 of these plus the 4 sides of the quadrilateral, summing to 340 / 3+66 = 538 / 3. Thus a + b = 541 . Interesting alternate way to calculate the circumradius and the area of the quadrilateral: We can also note that the circumradius and the area of a quadrilateral does not change if we swap two of its side lengths. Therefore we can consider the cyclic quadrilateral with side lengths 7 , 24 , 15 , 20, and note that one of its diagonals is the hypoteneuse of two pythagorean triples (This can be shown via the law of cosines on the two triangles formed by this diagonal, and noting the cosines of opposite vertices (which are supplementary) have the same value but opposite sign.), so it has length 25, and the two vertices opposite it are right angles. Thus the diagonal is a diameter, and the circumradius is half of that, or 25 / 2. The area can also be computed as the sum of the areas of the two right triangles, which is 84 + 150 = 234.