HMMT 二月 2022 · 几何 · 第 7 题
HMMT February 2022 — Geometry — Problem 7
题目详情
- Point P is located inside a square ABCD of side length 10. Let O , O , O , O be the circumcenters 1 2 3 4 √ of P AB , P BC , P CD , and P DA , respectively. Given that P A + P B + P C + P D = 23 2 and the area p a of O O O O is 50, the second largest of the lengths O O , O O , O O , O O can be written as , 1 2 3 4 1 2 2 3 3 4 4 1 b where a and b are relatively prime positive integers. Compute 100 a + b .
解析
- Point P is located inside a square ABCD of side length 10. Let O , O , O , O be the circumcenters 1 2 3 4 √ of P AB , P BC , P CD , and P DA , respectively. Given that P A + P B + P C + P D = 23 2 and the area p a of O O O O is 50, the second largest of the lengths O O , O O , O O , O O can be written as , 1 2 3 4 1 2 2 3 3 4 4 1 b where a and b are relatively prime positive integers. Compute 100 a + b . Proposed by: Daniel Zhu Answer: 16902 Solution: Note that O O and O O are perpendicular and intersect at O, the center of square ABCD . 1 3 2 4 Also note that O O , O O , O O , O O are the perpendiculars of P B, P C, P D, P A, respectively. Let 1 2 2 3 3 4 4 1 d = OO , d = OO , d = OO , and d = OO . Note that that since the area of O O O O = 50 , 1 1 2 2 3 3 4 4 1 2 3 4 we have that ( d + d )( d + d ) = 100. Also note that the area of octagon AO BO CO DO is twice 1 3 2 4 1 2 3 4 the area of O O O O , which is the same as the area of ABCD . Note that the difference between the 1 2 3 4 1 area of this octagon and ABCD is · 10[( d − 5) + ( d − 5) + ( d − 5) + ( d − 5)]. Since this must equal 1 2 3 4 2 0, we have d + d + d + d = 20. Combining this with the fact that ( d + d )( d + d ) = 100 gives us 1 2 3 4 1 3 2 4 d + d = d + d = 10, so O O = O O = 10. Note that if we translate AB by 10 to coincide with 1 3 2 4 1 3 2 4 ′ ′ DC , then O would coincide with O , and thus if P translates to P , then P CP D is cyclic. In other 1 3 words, we have ∠ AP B and ∠ CP D are supplementary. ◦ ◦ Fix any α ∈ (0 , 180 ). There are at most two points P in ABCD such that ∠ AP B = α and ◦ ′ ∠ CP D = 180 − α (two circular arcs intersect at most twice). Let P denote the unique point on AC ′ ∗ ∗ such that ∠ AP B = α , and let P denote the unique point on BD such that ∠ AP B = α . Note that ′ ∗ ◦ ′ ∗ it is not hard to see that in we have ∠ CP D = ∠ CP D = 180 − α . Thus, we have P = P or P = P , ′ so P must lie on one of the diagonals. Without loss of generality, assume P = P ( P is on AC ). Note that O O O O is an isosceles trapezoid with bases O O and O O . Additionally, the height of 1 2 3 4 1 4 2 3 √ AC the trapezoid is = 5 2. Since the area of trapezoid is O O O O , we have the midlength of the 1 2 3 4 2 √ 50 ◦ ◦ √ trapezoid is = 5 2. Additionally, note that ∠ P O B = 2 ∠ P AB = 90 . Similarly ∠ P O B = 90 . 1 2 5 2 Combining this with the fact that O O perpendicular bisects P B , we get that P O BO is a square, 1 2 1 2 q √ √ √ 23 2 − 10 2 13 2 169 so O O = P B = = = . Since this is the second largest side of O O O O , we 1 2 1 2 3 4 2 2 2 are done.