1991 AIME Problems/Problem 12

Revision as of 22:36, 27 August 2023 by Yiyj1 (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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 $m+n^{}_{}$.


[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 $\frac{672}{5}\implies\boxed{677}$

See also

1991 AIME (ProblemsAnswer KeyResources)
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. AMC logo.png