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>m/n^{}_{}</math>, in lowest terms, denote the [[perimeter]] of <math>ABCD^{}_{}</math>. Find <math>m+n^{}_{}</math>.
+[[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 ==
-{{image}}
+<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.
+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>.
@@ Line 16: / Line 16: @@
 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 [[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 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> similarity. From here, it's clear that
+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 {IP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}.
+\frac {AD}{BD} = \frac {OP}{PQ}\implies \frac {AD}{48} = \frac {15}{25}\implies AD = \frac {144}{5}.
 </cmath>
 Similarly,
@@ Line 30: / Line 30: @@
 === 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>:
-<math>BP \cdot PQ = \frac {OB}{2} \cdot PQ \Rightarrow OB = \frac {2BP\cdot PQ}{OB} \Rightarrow OB = 24</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>
-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 - \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>.
-From the right triangle <math>CRQ</math> we have <math>RC^2 = 25^2 - (\frac {44}{5})^2\Rightarrow RC = \frac {117}{5}</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 we can define a similar formula : <math>OB^2 = BP \cdot BA</math> , and then we find <math>AP</math>
+=== Solution 4 ===
-in other words the segment <math>OB</math> is tangent to the circles with diameters <math>AO,CO)</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>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>.
-The perimeter is <math>2(PB + BQ + QC + CR) = 2(15 + 20 + \frac {44 + 117}{5}) = \frac {672}{5}\Rightarrow m+n=677</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>.
-=== Solution 4 ===
+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>.
-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 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.
-=== Solution 5 ===
+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.
-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)
+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>.
-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)
+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>.
-So the perimeter is equal to <math>(38.4 + 28.8)*2</math> which equals <math>\frac {672}{5}</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 ==
@@ Line 64: / Line 71: @@
 [[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

1991 AIME (Problems • Answer Key • Resources)
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