Question

In $\triangle ABC,$ $BC = 23,$ $CA = 27,$ and $AB = 30.$ Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW},$ points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY},$ and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}.$ In addition, the points are positioned so that $\overline{UV} \parallel \overline{BC},$ $\overline{WX} \parallel \overline{AB},$ and $\overline{YZ} \parallel \overline{CA}.$ Right angle folds are then made along $\overline{UV},$ $\overline{WX},$ and $\overline{YZ}.$ The resulting figure is placed on a level floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from $\triangle ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\frac{k\sqrt{m}}{n},$ where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k + m + n.$

答案

Solution:

Write $a = BC = 23,$ $b = CA = 27,$ $c = AB = 30,$ and let $K$ be the area of $\triangle ABC.$ By Heron's formula with semiperimeter $40,$ $K = \sqrt{40 \cdot 17 \cdot 13 \cdot 10} = 20\sqrt{221}.$ When the corner at a vertex is folded down at a right angle, the flap hangs to a depth equal to the distance from that vertex to the fold line, so for a level tabletop of height $h,$ each fold line must lie at distance $h$ from its vertex.

The flap at $A$ is similar to $\triangle ABC$ with ratio $\frac{h}{2K/a} = \frac{ha}{2K}$ (dividing $h$ by the distance from $A$ to $\overline{BC}$ ), so it uses up $AU = c \cdot \frac{ha}{2K}$ of side $\overline{AB};$ likewise the flap at $B$ uses $BZ = c \cdot \frac{hb}{2K}$ of the same side. The two folds fit without crossing exactly when $AU + BZ \le c,$ that is, $h(a + b) \le 2K.$ The other two sides give $h(b + c) \le 2K$ and $h(c + a) \le 2K.$

The binding constraint comes from the largest sum, $b + c = 57,$ so the maximum height is $h = \frac{2K}{57} = \frac{40\sqrt{221}}{57},$ and $k + m + n = 40 + 221 + 57 = 318.$

2011 AIME I Problem 8

Solution:

Problem 8 in Other Years