Difference between revisions of "1991 AIME Problems/Problem 12"
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=== Solution 2 === | === Solution 2 === | ||
− | From above, we have <math>OB = 24</math> and <math>BD = 48</math>. Returning to <math>BPQO,</math> note that <math>\angle PQO\cong \angle PBO \cong ABD.</math> Hence, <math>\triangle ABD \sim \triangle OQP</math> by <math>AA</math> | + | From above, we have <math>OB = 24</math> and <math>BD = 48</math>. Returning to <math>BPQO,</math> note that <math>\angle PQO\cong \angle PBO \cong ABD.</math> Hence, <math>\triangle ABD \sim \triangle OQP</math> by <math>AA</math> [[similar triangle|similar]]ity. From here, it's clear that |
<cmath> | <cmath> | ||
\frac {AD}{BD} = \frac {IP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}. | \frac {AD}{BD} = \frac {IP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}. | ||
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=== Solution 3 === | === Solution 3 === | ||
− | The triangles <math>QOB,OBC</math> are [[isosceles triangle|isosceles]], and similar (because they have <math>\angle QOB = \angle OBC</math>). | + | The triangles <math>QOB,OBC</math> are [[isosceles triangle|isosceles]], and [[similar triangles|similar]] (because they have <math>\angle QOB = \angle OBC</math>). |
Hence <math>\frac {BQ}{OB} = \frac {OB}{BC} \Rightarrow OB^2 = BC \cdot BQ</math> | Hence <math>\frac {BQ}{OB} = \frac {OB}{BC} \Rightarrow OB^2 = BC \cdot BQ</math> | ||
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=== Solution 4 === | === Solution 4 === | ||
− | For convenience, let <math>\angle PQS = \theta</math>. Since the opposite triangles are congruent we have that <math>\angle BQR = 3\theta</math>, and therefore <math>\angle QRC = 3\theta - 90</math>. Let <math>RC = a</math>, then we have <math>\sin{(3\theta - 90)} = \frac {a}{25}</math>, or <math>- \cos{3\theta} = \frac {a}{25}</math>. Expanding with the formula <math>\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}</math>, and since we have <math>\cos{\theta} = \frac {4}{5}</math>, | + | For convenience, let <math>\angle PQS = \theta</math>. Since the opposite triangles are congruent we have that <math>\angle BQR = 3\theta</math>, and therefore <math>\angle QRC = 3\theta - 90</math>. Let <math>RC = a</math>, then we have <math>\sin{(3\theta - 90)} = \frac {a}{25}</math>, or <math>- \cos{3\theta} = \frac {a}{25}</math>. Expanding with the formula <math>\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}</math>, and since we have <math>\cos{\theta} = \frac {4}{5}</math>, we can solve for <math>a</math>. The rest then follows similarily from above. |
=== Solution 5 === | === Solution 5 === | ||
− | You can just | + | You can just label the points. After drawing a brief picture, you can see 4 right triangles with sides of <math>15,\ 20,\ 25</math>. |
− | + | Let the points of triangle <math>RDS</math> be <math>(0,0)\ (0,20)\ (15,0)</math>. Since each right triangle can be split into two similar triangles, point <math>(0,0)</math> is <math>12</math> away from the [[hypotenuse]]. By reflecting <math>(0,0)</math> 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 <math>(19.2,14.4)</math>. | |
− | Since each right triangle can be split into two similar triangles, point (0,0) is 12 away from the | ||
− | By reflecting (15,0) over diagonal SQ we get P (23.4,28.8). | + | By reflecting <math>(15,0)</math> over diagonal <math>\overline{SQ}</math> we get <math>P (23.4,28.8)</math>. By adding <math>15</math> to the <math>x</math> value we get <math>B(38.4,28.8)</math>. |
− | So the perimeter is equal to <math>(38.4 + 28.8)*2 | + | So the perimeter is equal to <math>(38.4 + 28.8)*2 = \frac {672}{5}</math>. |
== See also == | == See also == |
Revision as of 17:12, 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 .
Contents
[hide]Solution
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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 :
So
From the right triangle we have
or we can define a similar formula : , and then we find in other words 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 |