Difference between revisions of "1991 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | [[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 | + | [[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>\frac{m}{n}</math>, in lowest terms, denote the [[perimeter]] of <math>ABCD^{}_{}</math>. Find <math>m+n^{}_{}</math>. |
__TOC__ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | <center><asy>defaultpen(fontsize(12)+linewidth(1.3)); 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(" | |
=== Solution 1 === | === Solution 1 === | ||
− | Let <math>O</math> be the center of the rhombus. Via [[parallel]] sides and | + | 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. |
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 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>. | ||
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x = \frac{192 \pm \sqrt{192^2 - 4 \cdot 5 \cdot 1755}}{10} &=& 15, \frac{117}{5}\end{eqnarray*}</cmath> | 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 | + | We reject <math>15</math> because then everything degenerates into squares, 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 === | === 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 { | + | \frac {AD}{BD} = \frac {OP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}. |
</cmath> | </cmath> | ||
Similarly, | Similarly, | ||
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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>. |
The length of <math>OB</math> could be found easily from the area of <math>BPQ</math>: | The length of <math>OB</math> could be found easily from the area of <math>BPQ</math>: | ||
− | < | + | <cmath>BP \cdot BQ = \frac {OB}{2} \cdot PQ \Rightarrow OB = \frac {2BP\cdot BQ}{PQ} \Rightarrow OB = 24</cmath> |
+ | <cmath>OB^2 = BC \cdot BQ \Rightarrow 24^2 = (20 + CQ) \cdot 20 \Rightarrow CQ = \frac {44}{5}</cmath> | ||
− | + | From the right triangle <math>CRQ</math> we have <math>RC^2 = 25^2 - \left(\frac {44}{5}\right)^2\Rightarrow RC = \frac {117}{5}</math>. We could have also defined a similar formula: <math>OB^2 = BP \cdot BA</math>, and then we found <math>AP</math>, 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\left(15 + 20 + \frac {44 + 117}{5}\right) = \frac {672}{5}\Rightarrow m+n=677</math>. | |
− | <math>(</math>or | + | === 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>QC = 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 === | ||
+ | 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 point <math>E</math> be on <math>\overline{SR}</math>, such that <math>SE = 16</math> and <math>ER = 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 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>. |
− | + | ||
+ | === Solution 6 === | ||
+ | We can just use areas. Let <math>AP = b</math> and <math>AS = a</math>. <math>a^2 + b^2 = 625</math>. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, <math>(a+20)(b+15)</math>. This gives <math>3a + 4b = 120</math>. Solving this system of equation gives <math>\frac{44}{5} = a</math>, <math>\frac{117}{5} = b</math>, from which it is straightforward to find the answer, <math>2(a+b+35) \Rightarrow \frac{672}{5}</math>. Thus, <math>m+n = \frac{672}{5}\implies\boxed{677}</math> | ||
+ | |||
+ | === Solution 7 === | ||
+ | We will bash with trigonometry. | ||
− | === | + | Firstly, by Pythagoras Theorem, <math>PQ=QR=RS=SP=25</math>. We observe that <math>[PQRS]=\frac{1}{2}\cdot30\cdot40=600</math>. Thus, if we drop an altitude from <math>P</math> to <math>\overline{SR}</math> to point <math>E</math>, it will have length <math>\frac{600}{25}=24</math>. In particular, <math>SE=7</math> since we form a 7-24-25 triangle. |
− | |||
− | + | Now, <math>\sin\angle APS=\sin\angle SPB=\sin(\angle SPQ+\angle QPB)=\sin\angle SPQ\cos\angle QPB+\sin\angle QPB\cos\angle SPQ=\sin\angle PSR\cos\angle QPB-\sin\angle QPB\cos\angle PSR=\frac{24}{25}\cdot\frac{15}{25}-\frac{20}{25}\cdot\frac{7}{25}=\frac{44}{125}</math>. Thus, since <math>PS=25</math>, we get that <math>AS=\frac{44}{5}</math>. Now, by the Pythagorean Theorem, <math>AP=\frac{117}{5}</math>. | |
− | |||
− | + | Using the same idea, <math>\cos\angle RSD=-\cos\angle RSA=-\cos(\angle RSP+\angle PSA)=\sin\angle RSP\sin\angle PSA-\cos\angle RSP\cos\angle PSA=\frac{24}{25}\cdot\frac{117}{125}-\frac{7}{25}\cdot\frac{44}{125}=\frac{4}{5}</math>. Thus, since <math>SR=20</math>. | |
− | + | Now, we can finish. We know <math>AB=\frac{117}{5}+15=\frac{192}{5}</math>. We also know <math>AD=\frac{44}{5}+20=\frac{144}{5}</math>. Thus, our perimeter is <math>\frac{672}{5}\implies\boxed{677}</math> | |
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:36, 27 August 2023
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
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 .
Solution 6
We can just use areas. Let and . . Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, . This gives . Solving this system of equation gives , , from which it is straightforward to find the answer, . Thus,
Solution 7
We will bash with trigonometry.
Firstly, by Pythagoras Theorem, . We observe that . Thus, if we drop an altitude from to to point , it will have length . In particular, since we form a 7-24-25 triangle.
Now, . Thus, since , we get that . Now, by the Pythagorean Theorem, .
Using the same idea, . Thus, since .
Now, we can finish. We know . We also know . Thus, our perimeter is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.