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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $BPQ$ :

$BP \cdot PQ = \frac {OB}{2} \cdot PQ \Rightarrow OB = \frac {2BP\cdot PQ}{OB} \Rightarrow OB = 24$

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

From the right triangle $CRQ$ we have $RC^2 = 25^2 - (\frac {44}{5})^2\Rightarrow RC = \frac {117}{5}$

$($ or we can define a similar formula : $OB^2 = BP \cdot BA$ , and then we find $AP$ in other words the segment $OB$ is tangent to the circles with diameters $AO,CO)$

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

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 $\cos{\theta} = \frac {4}{5}$ , the rest follows similarily to above.

Solution 5

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

So the points of triangle RDS are (0,0) (0,20) and (15,0) Since each right triangle can be split into two similar triangles, point (0,0) is 12 away from the hypotnuse. By reflecting (0,0) over the hypotnuse, we can get 3rd point of the second right triangle (A.K.A the intersection of the diagonals of the Rhombus) which is (19.2,14.4)

By reflecting (15,0) over diagonal 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 $(38.4 + 28.8)*2$ which equals $\frac {672}{5}$ .

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
All AIME Problems and Solutions

Page

Toolbox

Search