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Difference between revisions of "1991 AIME Problems/Problem 12"

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== Solution ==
 
== Solution ==
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Let <math>O</math> be the center of the rhombus. Via [[parallel]] sides and [[alternate interior angles]], we see that the opposite [[triangle]]s are [[congruent]] (<math>\triangle BPQ \cong \triangle DRS</math>, <math>\triangle APS \cong CRQ</math>). Quickly we realize that <math>O</math> is also the center of the rectangle.  
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Let <math>O</math> be the center of the rhombus. Via [[parallel]] sides and [[alternate interior angles]], we see that the opposite [[triangle]]s are [[congruent]] (<math>\triangle BPQ \cong \triangle DRS</math>, <math>\triangle APS \cong \triangle CRQ</math>). Quickly we realize that <math>O</math> is also the center of the rectangle.  
  
 
By the [[Pythagorean Theorem]], we can solve for a side of the rhombus; <math>PQ = \sqrt{15^2 + 20^2} = 25</math>. Since the [[diagonal]]s of a rhombus are [[perpendicular bisector]]s, we have that <math>OP = 15, OQ = 20</math>. Also, <math>\angle POQ = 90^{\circ}</math>, so quadrilateral <math>BPOQ</math> is [[cyclic quadrilateral|cyclic]]. By [[Ptolemy's Theorem]], <math>25 \cdot OB = 20 \cdot 15 + 15 \cdot 20 = 600</math>.
 
By the [[Pythagorean Theorem]], we can solve for a side of the rhombus; <math>PQ = \sqrt{15^2 + 20^2} = 25</math>. Since the [[diagonal]]s of a rhombus are [[perpendicular bisector]]s, we have that <math>OP = 15, OQ = 20</math>. Also, <math>\angle POQ = 90^{\circ}</math>, so quadrilateral <math>BPOQ</math> is [[cyclic quadrilateral|cyclic]]. By [[Ptolemy's Theorem]], <math>25 \cdot OB = 20 \cdot 15 + 15 \cdot 20 = 600</math>.

Revision as of 16:55, 23 October 2007

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.

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 $m + n = \boxed{677}$.

See also

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
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