Difference between revisions of "1991 AIME Problems/Problem 12"
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=== Solution 5 === | === Solution 5 === | ||
− | + | We can just find coordinates of the points. After drawing a picture, we can see 4 congruent right triangles with sides of <math>15,\ 20,\ 25</math>, namely triangles <math>DSR, OSR, OQP,</math> and <math>BQP</math>. | |
− | Let the points of triangle <math>DSR</math> be <math>(0,0)\ (0,20)\ (15,0)</math>. | + | Let the points of triangle <math>DSR</math> be <math>(0,0)\ (0,20)\ (15,0)</math>. Let point <math>E</math> be on <math>\overline{SR}</math>, such that <math>SE = 16</math> and <math>SR = 9</math>. Triangle <math>DSR</math> can be split into two similar 3-4-5 right triangles, <math>ESD</math> and <math>EDR</math>. By the Pythagorean Theorem, point <math>D</math> is <math>12</math> away from point <math>E</math>. Repeating the process, if we break down triangle <math>DER</math> into two more similar triangles, we find that point <math>E</math> is at <math>(9.6, 7.2)</math>. |
− | By reflecting <math>( | + | By reflecting point <math>D = (0,0)</math> over point <math>E = (9.6, 7.2)</math>, we get point <math>O = (19.2, 14.4)</math>. By reflecting point <math>D</math> over point <math>O</math>, we get point <math>B = (38.4, 28.8)</math>. Thus, the perimeter is equal to <math>(38.4 + 28.8)\times 2 = \frac {672}{5}</math>, making the final answer <math>672+5 = 677</math>. |
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== See also == | == See also == |
Revision as of 18:49, 21 August 2010
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
![[asy]defaultpen(fontsize(10)+linewidth(0.65)); pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P=(23.4,28.8), Q=(38.4,8.8), R=(15,0), S=(0,20); O=intersectionpoint(A--C,B--D); draw(A--B--C--D--cycle);draw(P--R..Q--S); draw(P--Q--R--S--cycle); label("\(A\)",A,NW);label("\(B\)",B,NE);label("\(C\)",C,SE);label("\(D\)",D,SW); label("\(P\)",P,N);label("\(Q\)",Q,E);label("\(R\)",R,SW);label("\(S\)",S,W); label("\(15\)",B/2+P/2,N);label("\(20\)",B/2+Q/2,E);label("\(O\)",O,SW); [/asy]](http://latex.artofproblemsolving.com/b/a/5/ba5d2053ed60d0f7642922cfda4027d7224d527a.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
We can just find coordinates of the points. After drawing a picture, we can see 4 congruent right triangles with sides of , namely triangles
and
.
Let the points of triangle be
. Let point
be on
, such that
and
. Triangle
can be split into two similar 3-4-5 right triangles,
and
. By the Pythagorean Theorem, point
is
away from point
. Repeating the process, if we break down triangle
into two more similar triangles, we find that point
is at
.
By reflecting point over point
, we get point
. By reflecting point
over point
, we get point
. Thus, the perimeter is equal to
, making the final answer
.
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 |