Difference between revisions of "1991 AIME Problems/Problem 12"
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== Solution == | == Solution == | ||
− | {{ | + | 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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+ | 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>. | ||
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+ | By similar logic, we have <math>APOS</math> is a cyclic quadrilateral. Let <math>AP = x</math>, <math>AS = y</math>. The Pythagorean Theorem gives us <math>x^2 + y^2 = 625\quad \mathrm{(1)}</math>. Ptolemy’s Theorem gives us <math>25 \cdot OA = 20x + 15y</math>. Since the diagonals of a rectangle are equal, <math>OA = \frac{1}{2}d = OB</math>, and <math>20x + 15y = 600\quad \mathrm{(2)}</math>. Solving for <math>y</math>, we get <math>y = 40 - \frac 43x</math>. Substituting into <math>\mathrm{(1)}</math>, | ||
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+ | <cmath>\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*}</cmath> | ||
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+ | We reject <math>15</math> because then everything degenerates into [[square]]s, but the condition that <math>PR \neq QS</math> gives us a [[contradiction]]. Thus <math>x = \frac{117}{5}</math>, and backwards solving gives <math>y = \frac{44}5</math>. The perimeter of <math>ABCD</math> is <math>2\left(20 + 15 + \frac{117}{5} + \frac{44}5\right) = \frac{672}{5}</math>, and <math>m + n = \boxed{677}</math>. | ||
== See also == | == See also == |
Revision as of 16:55, 23 October 2007
Problem
Rhombus is inscribed in rectangle so that vertices , , , and are interior points on sides , , , and , respectively. It is given that , , , and . Let , in lowest terms, denote the perimeter of . Find .
Solution
Let be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent (, ). Quickly we realize that is also the center of the rectangle.
By the Pythagorean Theorem, we can solve for a side of the rhombus; . Since the diagonals of a rhombus are perpendicular bisectors, we have that . Also, , so quadrilateral is cyclic. By Ptolemy's Theorem, .
By similar logic, we have is a cyclic quadrilateral. Let , . The Pythagorean Theorem gives us . Ptolemy’s Theorem gives us . Since the diagonals of a rectangle are equal, , and . Solving for , we get . Substituting into ,
We reject because then everything degenerates into squares, but the condition that gives us a contradiction. Thus , and backwards solving gives . The perimeter of is , and .
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 |