AIME 2026 II · 第 3 题

AIME 2026 II — Problem 3

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Let $ABCDE$ be a nonconvex pentagon with internal angles $\angle A = \angle E = 90^\circ$ and $\angle B = \angle D = 45^\circ$ . Suppose that $DE < AB$ , $AE = 20$ , $BC = 14\sqrt2$ , and points $B$ , $C$ , and $D$ lie on the same side of line $AE$ . Suppose further that $AB$ is an integer with $AB < 2026$ and the area of pentagon $ABCDE$ is an integer multiple of $16$ . Find the number of possible values of $AB$ .

$AIME diagram$

~Diagram by sillybone

解析

Solution 1

Construct line $FG$ such that it passes through point $C$ and is parallel to line $AE$ . Since $FG||AE$ , $\angle BFC = \angle A = 90^\circ.$ Since $\angle B = 45^\circ$ , triangle $\triangle BFC$ is a $45-45-90$ triangle, meaning that $FC=14,$ and $CG=6.$ Since $\angle B = 45^\circ$ and $\angle DGC=90^\circ$ from parallel lines, triangle $\triangle DCG \sim{\triangle BCF}.$ Therefore, $BF=FC=14$ , and $CG=GD=6.$ If we set the length of segment $AF$ to $x$ , we can get the area of pentagon $ABCDE$ as $116+20x.$ Since $116+20x$ must be a multiple of $16, x\equiv 3\mod{4}.$ Since $x+14<2026, x<2012.$ We can express $x$ as $4y+3,$ so therefore $4y+3<2012,$ so $4y<2009.$ Because $y>=0,$ solving the first inequality gives $y<502\frac{1}{4},$ meaning that $0\le y \le 502.$ Since $y$ must be an integer, there are $503$ values of $y,$ giving $\boxed{503}$ possible values of $AB.$

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