Difference between revisions of "1991 AIME Problems/Problem 12"
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== Solution == | == Solution == | ||
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=== Solution 1 === | === Solution 1 === | ||
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. | 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. |
Revision as of 19:11, 3 November 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
.
Contents
Solution
![AIME 1991 Solution 12.png](https://wiki-images.artofproblemsolving.com//0/0a/AIME_1991_Solution_12.png)
Solution 1
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
.
Solution 2
From above, we have and
. Returning to
note that
Hence,
by
similarity. From here, it's clear that
Similarly,
Therefore, the perimeter of rectangle
is
Solution 3
The triangles are isosceles, and similar (because they have
).
Hence .
The length of could be found easily from the area of
:
From the right triangle we have
. We could have also defined a similar formula:
, and then we found
, the segment
is tangent to the circles with diameters
.
The perimeter is .
Solution 4
For convenience, let . Since the opposite triangles are congruent we have that
, and therefore
. Let
, then we have
, or
. Expanding with the formula
, and since we have
, we can solve for
. 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 .
Let the points of triangle be
. Since each right triangle can be split into two similar triangles, point
is
away from the hypotenuse. By reflecting
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
.
By reflecting over diagonal
we get
. By adding
to the
value we get
.
So the perimeter is equal to .
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 |