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. |
+ | |||
+ | 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 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>, | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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 17:55, 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 |