Question

Let $u$ and $v$ be integers satisfying $0 \lt v \lt u.$ Let $A = (u, v),$ let $B$ be the reflection of $A$ across the line $y = x,$ let $C$ be the reflection of $B$ across the $y$ -axis, let $D$ be the reflection of $C$ across the $x$ -axis, and let $E$ be the reflection of $D$ across the $y$ -axis. The area of pentagon $ABCDE$ is $451.$ Find $u + v.$

答案

Solution:

Carrying out the reflections, $B = (v, u),$ $C = (-v, u),$ $D = (-v, -u),$ and $E = (v, -u).$ The points $B, C, D, E$ form a rectangle of width $2v$ and height $2u,$ with area $4uv,$ and $A = (u, v)$ sticks out to its right. Triangle $ABE$ has vertical base $BE$ of length $2u$ and horizontal height $u - v,$ so its area is $u(u - v).$

The pentagon's area is therefore $4uv + u(u - v) = u^2 + 3uv = u(u + 3v) = 451 = 11 \cdot 41.$ Since $0 \lt v \lt u,$ we have $u \lt u + 3v \lt 4u,$ which rules out the factorization $1 \cdot 451.$ So $u = 11$ and $u + 3v = 41,$ giving $v = 10,$ which indeed satisfies $v \lt u.$

Thus $u + v = 11 + 10 = 21.$

2000 AIME I Problem 2

Solution:

Problem 2 in Other Years