1991 AIME Problem 12

Attempt Problem 12 of the 1991 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AIME solutions, or check the answer key.

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12.

Rhombus PQRSPQRS is inscribed in rectangle ABCDABCD so that vertices P,P, Q,Q, R,R, and SS are interior points on sides AB,\overline{AB}, BC,\overline{BC}, CD,\overline{CD}, and DA,\overline{DA}, respectively. It is given that PB=15,PB=15, BQ=20,BQ=20, PR=30,PR=30, and QS=40.QS=40. Let mn,\frac{m}{n}, in lowest terms, denote the perimeter of ABCD.ABCD. Find m+n.m+n.

Answer: 677
Concepts:rhombuscoordinate geometryvector
Difficulty rating: 2350
Small Hint:

The center of the rhombus is also the center of the rectangle, and its half-diagonals have lengths 1515 and 2020

Big Hint:

Use coordinates for PP and QQ; their vectors from the common center are perpendicular

Solution:

Let the rectangle have width ww and height h,h, with A=(0,0)A=(0,0) and B=(w,0).B=(w,0). Then P=(w15,0)P=(w-15,0) and Q=(w,20).Q=(w,20). The diagonals of the rhombus bisect each other at the rectangle’s center O=(w2,h2).O=(\frac{w}{2},\frac{h}{2}). Put p=OP=(w215,h2).p=\overrightarrow{OP}=(\frac{w}{2}-15,-\frac{h}{2}). Since OP=15,OP=15, OQ=20,OQ=20, and the rhombus diagonals are perpendicular, while PQ=(15,20),\overrightarrow{PQ}=(15,20), we have p=15,p(15,20)=225.\begin{aligned}|p|&=15,\\p\mathbin{\cdot}(15,20)&=-225.\end{aligned} Solving these two equations, with the second coordinate negative because h>0,h>0, gives p=(215,725).p=(\frac{21}{5},-\frac{72}{5}). Hence w=2(15+215)=1925,h=1445.\begin{aligned}w&=2\left(15+\frac{21}{5}\right)=\frac{192}{5},\\h&=\frac{144}{5}.\end{aligned} The perimeter is 2(w+h)=6725,2(w+h)=\frac{672}{5}, so m+n=672+5=677.m+n=672+5=677.

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