Difference between revisions of "1991 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | Rhombus <math>PQRS^{}_{}</math> is | + | [[Rhombus]] <math>PQRS^{}_{}</math> is [[inscribe]]d in [[rectangle]] <math>ABCD^{}_{}</math> so that [[vertex|vertices]] <math>P^{}_{}</math>, <math>Q^{}_{}</math>, <math>R^{}_{}</math>, and <math>S^{}_{}</math> are interior points on sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math>, and <math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_{}=30</math>, and <math>QS^{}_{}=40</math>. Let <math>m/n^{}_{}</math>, in lowest terms, denote the [[perimeter]] of <math>ABCD^{}_{}</math>. Find <math>m+n^{}_{}</math>. |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
{{image}} | {{image}} | ||
− | + | === 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. | ||
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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>. | 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>. | ||
+ | |||
+ | === 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> similarity. From here, it's clear that | ||
+ | <cmath> | ||
+ | \frac {AD}{BD} = \frac {IP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}. | ||
+ | </cmath> | ||
+ | Similarly, | ||
+ | <cmath> | ||
+ | \frac {AB}{BD} = \frac {IQ}{PQ}\implies \frac {AB}{48} = \frac {20}{25}\implies AB = \frac {192}{5}. | ||
+ | </cmath> | ||
+ | Therefore, the perimeter of rectangle <math>ABCD</math> is <math>2(AB + AD) = 2\left(\frac {192}{5} + \frac {144}{5}\right) = \frac {672}{5}.</math> | ||
+ | |||
+ | === Solution 3 === | ||
+ | The triangles <math>QOB,OBC</math> are [[isosceles triangle|isosceles]], and 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> | ||
+ | |||
+ | The length of <math>OB</math> could be found easily from the area of <math>BPQ</math>: | ||
+ | |||
+ | <math>BP \cdot PQ = \frac {OB}{2} \cdot PQ \Rightarrow OB = \frac {2BP\cdot PQ}{OB} \Rightarrow OB = 24</math> | ||
+ | |||
+ | So <math>OB^2 = BC \cdot BQ \Rightarrow 24^2 = (20 + CQ) \cdot 20 \Rightarrow CQ = \frac {44}{5}</math> | ||
+ | |||
+ | From the right triangle <math>CRQ</math> we have <math>RC^2 = 25^2 - (\frac {44}{5})^2\Rightarrow RC = \frac {117}{5}</math> | ||
+ | |||
+ | <math>(</math>or we can define a similar formula : <math>OB^2 = BP \cdot BA</math> , and then we find <math>AP</math> | ||
+ | in other words the segment <math>OB</math> is tangent to the circles with diameters <math>AO,CO)</math> | ||
+ | |||
+ | The perimeter is <math>2(PB + BQ + QC + CR) = 2(15 + 20 + \frac {44 + 117}{5}) = \frac {672}{5}\Rightarrow m+n=677</math>. | ||
+ | |||
+ | === 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>, the rest follows similarily to above. | ||
+ | |||
+ | === Solution 5 === | ||
+ | You can just lable the points. After drawing a brief picture, you can see 4 right triangles with sides of 15,20,25. | ||
+ | |||
+ | So the points of triangle RDS are (0,0) (0,20) and (15,0) | ||
+ | Since each right triangle can be split into two similar triangles, point (0,0) is 12 away from the hypotnuse. By reflecting (0,0) over the hypotnuse, we can get 3rd point of the second right triangle (A.K.A the intersection of the diagonals of the Rhombus) which is (19.2,14.4) | ||
+ | |||
+ | By reflecting (15,0) over diagonal SQ we get P (23.4,28.8). By adding 15 to the x value we get B(38.4,28.8) | ||
+ | |||
+ | So the perimeter is equal to <math>(38.4 + 28.8)*2</math> which equals <math>\frac {672}{5}</math>. | ||
== See also == | == See also == |
Revision as of 17:08, 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 , the rest follows similarily to above.
Solution 5
You can just lable the points. After drawing a brief picture, you can see 4 right triangles with sides of 15,20,25.
So the points of triangle RDS are (0,0) (0,20) and (15,0) Since each right triangle can be split into two similar triangles, point (0,0) is 12 away from the hypotnuse. By reflecting (0,0) over the hypotnuse, we can get 3rd point of the second right triangle (A.K.A the intersection of the diagonals of the Rhombus) which is (19.2,14.4)
By reflecting (15,0) over diagonal SQ we get P (23.4,28.8). By adding 15 to the x value we get B(38.4,28.8)
So the perimeter is equal to which equals .
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 |