Difference between revisions of "1991 AIME Problems/Problem 12"
m (Blanked the page) (Tag: Blanking) |
m (→Solution) |
||
(5 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
+ | == 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>\frac{m}{n}</math>, in lowest terms, denote the [[perimeter]] of <math>ABCD^{}_{}</math>. Find <math>m+n^{}_{}</math>. | ||
+ | __TOC__ | ||
+ | == 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("\(A\)",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></center> | ||
+ | === 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. | ||
+ | |||
+ | 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 similar logic, we have <math>APOS</math> is a cyclic quadrilateral. Let <math>AP = x</math>, <math>AS = y</math>. The Pythagorean Theorem gives us <math>x^2 + y^2 = 625\quad \mathrm{(1)}</math>. Ptolemy’s Theorem gives us <math>25 \cdot OA = 20x + 15y</math>. Since the diagonals of a rectangle are equal, <math>OA = \frac{1}{2}d = OB</math>, and <math>20x + 15y = 600\quad \mathrm{(2)}</math>. Solving for <math>y</math>, we get <math>y = 40 - \frac 43x</math>. Substituting into <math>\mathrm{(1)}</math>, | ||
+ | |||
+ | <cmath>\begin{eqnarray*}x^2 + \left(40-\frac 43x\right)^2 &=& 625\\ | ||
+ | 5x^2 - 192x + 1755 &=& 0\\ | ||
+ | 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 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 === | ||
+ | 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> | ||
+ | \frac {AD}{BD} = \frac {OP}{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 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | === 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 == | ||
+ | {{AIME box|year=1991|num-b=11|num-a=13}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:36, 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
Solution
![[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("\(A\)",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/0/a/b/0ab275747a2707450cd73d377348c4cc2543ee20.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
.
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.