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PUMaC 2014 · 几何(B 组) · 第 15 题

PUMaC 2014 — Geometry (Division B) — Problem 15

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

Geometry B

  1. [ 3 ] Triangle ABC has lengths AB = 20 , AC = 14 , BC = 22. The median from B intersects AC at M and the angle bisector from C intersects AB at N and the median from B at P . Let p [ AM P N ] = for positive integers p, q coprime. Note that [ ABC ] denotes the area of triangle q [ ABC ] ABC . Find p + q
  2. [ 3 ] Consider the pyramid OABC . Let the equilateral triangle ABC with side length 6 be the base. Also 9 = OA = OB = OC . Let M be the midpoint of AB . Find the square of the distance from M to OC .
  3. [ 4 ] As given in figure (not drawn to proportion), in 4 ABC , E ∈ AC , D ∈ AB , P = BE ∩ CD Given that S 4 BP C = 12, while the areas of 4 BP D , 4 CP E and quadrilateral AEP D are all the same, which is x . Find the value of x .
  4. [ 4 ] Let O be the circumcenter of triangle ABC with circumradius 15. Let G be the centroid o of ABC and let M be the midpoint of BC . If BC = 18 and ∠ M OA = 150 , find the area of OM G .
  5. [ 5 ] Consider the cyclic quadrilateral with sides 1, 4, 8, 7 in that order. What is its circumdi- √ ameter? Let the answer be of the form a b + c , for b square free. Find a + b + c
  6. [ 6 ] There is a point D on side AC of acute triangle 4 ABC . Let AM be the median drawn from A (so M is on BC ) and CH be the altitude drawn from C (so H is on AB ). Let I be the intersection of AM and CH , and let K be the intersection of AM and line segment BD . We know that AK = 8, BK = 8, and M K = 6. Find the length of AI .
  7. [ 7 ] Consider quadrilateral ABCD . Given that ∠ DAC = 70, ∠ BAC = 40, ∠ BDC = 20, ∠ CBD = 35. Let P be the intersection of AC and BD . Find ∠ BP C .
  8. [ 8 ] ABCD is a cyclic quadrilateral with circumcenter O and circumradius 7. AB intersects p CD at E , DA intersects CB at F . OE = 13, OF = 14. Let cos ∠ F OE = , with p, q coprime. q Find p + q . 1
解析

15 ) [ AOM ] = (12)(15) sin 150 = 45. Then G splits AM into the ratio 2 : 1 and so 2 GM 1 [ OM G ] = [ AOM ] = 45 = 15 . AM 3