Difference between revisions of "1991 AIME Problems/Problem 12"

Latest revision as of 22:36, 27 August 2023

Problem

Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$ , $Q^{}_{}$ , $R^{}_{}$ , and $S^{}_{}$ are interior points on sides $\overline{AB}$ , $\overline{BC}$ , $\overline{CD}$ , and $\overline{DA}$ , respectively. It is given that $PB^{}_{}=15$ , $BQ^{}_{}=20$ , $PR^{}_{}=30$ , and $QS^{}_{}=40$ . Let $\frac{m}{n}$ , in lowest terms, denote the perimeter of $ABCD^{}_{}$ . Find $m+n^{}_{}$ .

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]$

Solution 1

Let $O$ be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent ( $\triangle BPQ \cong \triangle DRS$ , $\triangle APS \cong \triangle CRQ$ ). Quickly we realize that $O$ is also the center of the rectangle.

By the Pythagorean Theorem, we can solve for a side of the rhombus; $PQ = \sqrt{15^2 + 20^2} = 25$ . Since the diagonals of a rhombus are perpendicular bisectors, we have that $OP = 15, OQ = 20$ . Also, $\angle POQ = 90^{\circ}$ , so quadrilateral $BPOQ$ is cyclic. By Ptolemy's Theorem, $25 \cdot OB = 20 \cdot 15 + 15 \cdot 20 = 600$ .

By similar logic, we have $APOS$ is a cyclic quadrilateral. Let $AP = x$ , $AS = y$ . The Pythagorean Theorem gives us $x^2 + y^2 = 625\quad \mathrm{(1)}$ . Ptolemy’s Theorem gives us $25 \cdot OA = 20x + 15y$ . Since the diagonals of a rectangle are equal, $OA = \frac{1}{2}d = OB$ , and $20x + 15y = 600\quad \mathrm{(2)}$ . Solving for $y$ , we get $y = 40 - \frac 43x$ . Substituting into $\mathrm{(1)}$ ,

$\begin{eqnarray*}x^2 + \left(40-\frac 43x\right)^2 &=& 625\\ 5x^2 - 192x + 1755 &=& 0\\ x = \frac{192 \pm \sqrt{192^2 - 4 \cdot 5 \cdot 1755}}{10} &=& 15, \frac{117}{5}\end{eqnarray*}$

We reject $15$ because then everything degenerates into squares, but the condition that $PR \neq QS$ gives us a contradiction. Thus $x = \frac{117}{5}$ , and backwards solving gives $y = \frac{44}5$ . The perimeter of $ABCD$ is $2\left(20 + 15 + \frac{117}{5} + \frac{44}5\right) = \frac{672}{5}$ , and $m + n = \boxed{677}$ .

Solution 2

From above, we have $OB = 24$ and $BD = 48$ . Returning to $BPQO,$ note that $\angle PQO\cong \angle PBO \cong ABD.$ Hence, $\triangle ABD \sim \triangle OQP$ by $AA$ similarity. From here, it's clear that $\frac {AD}{BD} = \frac {OP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}.$ Similarly, $\frac {AB}{BD} = \frac {IQ}{PQ}\implies \frac {AB}{48} = \frac {20}{25}\implies AB = \frac {192}{5}.$ Therefore, the perimeter of rectangle $ABCD$ is $2(AB + AD) = 2\left(\frac {192}{5} + \frac {144}{5}\right) = \frac {672}{5}.$

Solution 3

The triangles $QOB,OBC$ are isosceles, and similar (because they have $\angle QOB = \angle OBC$ ).

Hence $\frac {BQ}{OB} = \frac {OB}{BC} \Rightarrow OB^2 = BC \cdot BQ$ .

The length of $OB$ could be found easily from the area of $BPQ$ :

$BP \cdot BQ = \frac {OB}{2} \cdot PQ \Rightarrow OB = \frac {2BP\cdot BQ}{PQ} \Rightarrow OB = 24$ $OB^2 = BC \cdot BQ \Rightarrow 24^2 = (20 + CQ) \cdot 20 \Rightarrow CQ = \frac {44}{5}$

From the right triangle $CRQ$ we have $RC^2 = 25^2 - \left(\frac {44}{5}\right)^2\Rightarrow RC = \frac {117}{5}$ . We could have also defined a similar formula: $OB^2 = BP \cdot BA$ , and then we found $AP$ , the segment $OB$ is tangent to the circles with diameters $AO,CO$ .

The perimeter is $2(PB + BQ + QC + CR) = 2\left(15 + 20 + \frac {44 + 117}{5}\right) = \frac {672}{5}\Rightarrow m+n=677$ .

Solution 4

For convenience, let $\angle PQS = \theta$ . Since the opposite triangles are congruent we have that $\angle BQR = 3\theta$ , and therefore $\angle QRC = 3\theta - 90$ . Let $QC = a$ , then we have $\sin{(3\theta - 90)} = \frac {a}{25}$ , or $- \cos{3\theta} = \frac {a}{25}$ . Expanding with the formula $\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}$ , and since we have $\cos{\theta} = \frac {4}{5}$ , we can solve for $a$ . 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 $15,\ 20,\ 25$ , namely triangles $DSR, OSR, OQP,$ and $BQP$ .

Let the points of triangle $DSR$ be $(0,0)\ (0,20)\ (15,0)$ . Let point $E$ be on $\overline{SR}$ , such that $SE = 16$ and $ER = 9$ . Triangle $DSR$ can be split into two similar 3-4-5 right triangles, $ESD$ and $EDR$ . By the Pythagorean Theorem, point $D$ is $12$ away from point $E$ . Repeating the process, if we break down triangle $DER$ into two more similar triangles, we find that point $E$ is at $(9.6, 7.2)$ .

By reflecting point $D = (0,0)$ over point $E = (9.6, 7.2)$ , we get point $O = (19.2, 14.4)$ . By reflecting point $D$ over point $O$ , we get point $B = (38.4, 28.8)$ . Thus, the perimeter is equal to $(38.4 + 28.8)\times 2 = \frac {672}{5}$ , making the final answer $672+5 = 677$ .

Solution 6

We can just use areas. Let $AP = b$ and $AS = a$ . $a^2 + b^2 = 625$ . Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, $(a+20)(b+15)$ . This gives $3a + 4b = 120$ . Solving this system of equation gives $\frac{44}{5} = a$ , $\frac{117}{5} = b$ , from which it is straightforward to find the answer, $2(a+b+35) \Rightarrow \frac{672}{5}$ . Thus, $m+n = \frac{672}{5}\implies\boxed{677}$

Solution 7

We will bash with trigonometry.

Firstly, by Pythagoras Theorem, $PQ=QR=RS=SP=25$ . We observe that $[PQRS]=\frac{1}{2}\cdot30\cdot40=600$ . Thus, if we drop an altitude from $P$ to $\overline{SR}$ to point $E$ , it will have length $\frac{600}{25}=24$ . In particular, $SE=7$ since we form a 7-24-25 triangle.

Now, $\sin\angle APS=\sin\angle SPB=\sin(\angle SPQ+\angle QPB)=\sin\angle SPQ\cos\angle QPB+\sin\angle QPB\cos\angle SPQ=\sin\angle PSR\cos\angle QPB-\sin\angle QPB\cos\angle PSR=\frac{24}{25}\cdot\frac{15}{25}-\frac{20}{25}\cdot\frac{7}{25}=\frac{44}{125}$ . Thus, since $PS=25$ , we get that $AS=\frac{44}{5}$ . Now, by the Pythagorean Theorem, $AP=\frac{117}{5}$ .

Using the same idea, $\cos\angle RSD=-\cos\angle RSA=-\cos(\angle RSP+\angle PSA)=\sin\angle RSP\sin\angle PSA-\cos\angle RSP\cos\angle PSA=\frac{24}{25}\cdot\frac{117}{125}-\frac{7}{25}\cdot\frac{44}{125}=\frac{4}{5}$ . Thus, since $SR=20$ .

Now, we can finish. We know $AB=\frac{117}{5}+15=\frac{192}{5}$ . We also know $AD=\frac{44}{5}+20=\frac{144}{5}$ . Thus, our perimeter is $\frac{672}{5}\implies\boxed{677}$

@@ 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\\
+x^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 ==
-* [[1991 AIME Problems]]
+{{AIME box|year=1991|num-b=11|num-a=13}}
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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

Page

Toolbox

Search