1991 AIME Problems/Problem 12

Problem

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

Solution

$[asy]defaultpen(fontsize(10)+linewidth(0.65)); pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P=(23.4,28.8), Q=(38.4,8.8), R=(15,0), S=(0,20); O=intersectionpoint(A--C,B--D); draw(A--B--C--D--cycle);draw(P--R..Q--S); draw(P--Q--R--S--cycle); label("$A$",A,NW);label("$B$",B,NE);label("$C$",C,SE);label("$D$",D,SW); label("$P$",P,N);label("$Q$",Q,E);label("$R$",R,SW);label("$S$",S,W); label("$15$",B/2+P/2,N);label("$20$",B/2+Q/2,E);label("$O$",O,SW); [/asy]$

Solution 1

Let $O$ be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent ( $\triangle BPQ \cong \triangle DRS$ , $\triangle APS \cong \triangle CRQ$ ). Quickly we realize that $O$ is also the center of the rectangle.

By the Pythagorean Theorem, we can solve for a side of the rhombus; $PQ = \sqrt{15^2 + 20^2} = 25$ . Since the diagonals of a rhombus are perpendicular bisectors, we have that $OP = 15, OQ = 20$ . Also, $\angle POQ = 90^{\circ}$ , so quadrilateral $BPOQ$ is cyclic. By Ptolemy's Theorem, $25 \cdot OB = 20 \cdot 15 + 15 \cdot 20 = 600$ .

By similar logic, we have $APOS$ is a cyclic quadrilateral. Let $AP = x$ , $AS = y$ . The Pythagorean Theorem gives us $x^2 + y^2 = 625\quad \mathrm{(1)}$ . Ptolemy’s Theorem gives us $25 \cdot OA = 20x + 15y$ . Since the diagonals of a rectangle are equal, $OA = \frac{1}{2}d = OB$ , and $20x + 15y = 600\quad \mathrm{(2)}$ . Solving for $y$ , we get $y = 40 - \frac 43x$ . Substituting into $\mathrm{(1)}$ ,

$\begin{eqnarray*}x^2 + \left(40-\frac 43x\right)^2 &=& 625\\ 5x^2 - 192x + 1755 &=& 0\\ x = \frac{192 \pm \sqrt{192^2 - 4 \cdot 5 \cdot 1755}}{10} &=& 15, \frac{117}{5}\end{eqnarray*}$

We reject $15$ because then everything degenerates into squares, but the condition that $PR \neq QS$ gives us a contradiction. Thus $x = \frac{117}{5}$ , and backwards solving gives $y = \frac{44}5$ . The perimeter of $ABCD$ is $2\left(20 + 15 + \frac{117}{5} + \frac{44}5\right) = \frac{672}{5}$ , and $m + n = \boxed{677}$ .

Solution 2

From above, we have $OB = 24$ and $BD = 48$ . Returning to $BPQO,$ note that $\angle PQO\cong \angle PBO \cong ABD.$ Hence, $\triangle ABD \sim \triangle OQP$ by $AA$ similarity. From here, it's clear that $\frac {AD}{BD} = \frac {IP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}.$ Similarly, $\frac {AB}{BD} = \frac {IQ}{PQ}\implies \frac {AB}{48} = \frac {20}{25}\implies AB = \frac {192}{5}.$ Therefore, the perimeter of rectangle $ABCD$ is $2(AB + AD) = 2\left(\frac {192}{5} + \frac {144}{5}\right) = \frac {672}{5}.$

Solution 3

The triangles $QOB,OBC$ are isosceles, and similar (because they have $\angle QOB = \angle OBC$ ).

Hence $\frac {BQ}{OB} = \frac {OB}{BC} \Rightarrow OB^2 = BC \cdot BQ$ .

The length of $OB$ could be found easily from the area of $BPQ$ :

$BP \cdot PQ = \frac {OB}{2} \cdot PQ \Rightarrow OB = \frac {2BP\cdot PQ}{OB} \Rightarrow OB = 24$ $OB^2 = BC \cdot BQ \Rightarrow 24^2 = (20 + CQ) \cdot 20 \Rightarrow CQ = \frac {44}{5}$

From the right triangle $CRQ$ we have $RC^2 = 25^2 - \left(\frac {44}{5}\right)^2\Rightarrow RC = \frac {117}{5}$ . We could have also defined a similar formula: $OB^2 = BP \cdot BA$ , and then we found $AP$ , the segment $OB$ is tangent to the circles with diameters $AO,CO$ .

The perimeter is $2(PB + BQ + QC + CR) = 2\left(15 + 20 + \frac {44 + 117}{5}\right) = \frac {672}{5}\Rightarrow m+n=677$ .

Solution 4

For convenience, let $\angle PQS = \theta$ . Since the opposite triangles are congruent we have that $\angle BQR = 3\theta$ , and therefore $\angle QRC = 3\theta - 90$ . Let $RC = a$ , then we have $\sin{(3\theta - 90)} = \frac {a}{25}$ , or $- \cos{3\theta} = \frac {a}{25}$ . Expanding with the formula $\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}$ , and since we have $\cos{\theta} = \frac {4}{5}$ , we can solve for $a$ . The rest then follows similarily from above.

Solution 5

You can just label the points. After drawing a brief picture, you can see 4 right triangles with sides of $15,\ 20,\ 25$ .

Let the points of triangle $RDS$ be $(0,0)\ (0,20)\ (15,0)$ . Since each right triangle can be split into two similar triangles, point $(0,0)$ is $12$ away from the hypotenuse. By reflecting $(0,0)$ over the hypotenuse, we can get the 3rd point of the second right triangle (aka the intersection of the diagonals of the rhombus) which is $(19.2,14.4)$ .

By reflecting $(15,0)$ over diagonal $\overline{SQ}$ we get $P (23.4,28.8)$ . By adding $15$ to the $x$ value we get $B(38.4,28.8)$ .

So the perimeter is equal to $(38.4 + 28.8)*2 = \frac {672}{5}$ .

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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