Difference between revisions of "2023 AIME I Problems/Problem 8"

Revision as of 01:34, 10 February 2023

Problem

Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$

Diagram

$[asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T); draw(P--R^^P--S^^P--T,red+dashed); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(135),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot(R^^S^^T,linewidth(4.5)); dot(O,linewidth(4.5)); label("$9$",midpoint(P--R),dir(A-D),red); label("$5$",midpoint(P--S),dir(180),red); label("$16$",midpoint(P--T),dir(A-D),red); [/asy]$ ~MRENTHUSIASM

Solution 1

This solution refers to the Diagram section.

Let $O$ be the incenter of $ABCD$ for which $\odot O$ is tangent to $\overline{DA},\overline{AB},$ and $\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\overleftrightarrow{DA},\overleftrightarrow{AB},$ and $\overleftrightarrow{BC},$ respectively, such that $\overline{RT}$ intersects $\odot O$ at $P$ and $Q.$

We obtain the following diagram: $[asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; fill(R--T--Z--X--cycle,cyan); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T); draw(P--R^^P--S^^P--T,red+dashed); draw(O--X^^O--Y^^O--Z); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(135),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$S$",S,1.5*dir(-90),linewidth(4.5)); dot("$T$",T,1.5*dir(-45),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$X$",X,1.5*dir(135),linewidth(4.5)); dot("$Y$",Y,1.5*dir(-90),linewidth(4.5)); dot("$Z$",Z,1.5*dir(-45),linewidth(4.5)); dot("$Q$",Q,1.5*dir(60),linewidth(4.5)); label("$9$",midpoint(P--R),dir(A-D),red); label("$5$",midpoint(P--S),dir(180),red); label("$16$",midpoint(P--T),dir(A-D),red); [/asy]$ Note that $\angle RXZ = \angle TZX = 90^\circ$ by the properties of tangents, so $RTZX$ is a rectangle. It follows that the diameter of $\odot O$ is $XZ = RT = 25.$

Let $x=PQ$ and $y=RX=TZ.$ We apply the Power of a Point Theorem to $R$ and $T:$ $\begin{align*} y^2 &= 9(9+x), \\ y^2 &= 16(16-x). \end{align*}$ We solve this system of equations to get $x=7$ and $y=12.$ Alternatively, we can find these results by the symmetry on rectangle $RTZX$ and semicircle $\widehat{XPZ}.$

We extend $\overline{SP}$ beyond $P$ to intersect $\odot O$ and $\overleftrightarrow{CD}$ at $E$ and $F,$ respectively, where $E\neq P.$ So, we have $EF=SP=5$ and $PE=25-SP-EF=15.$ On the other hand, we have $PX=15$ by the Pythagorean Theorem on right $\triangle PRX.$ Together, we conclude that $E=X.$ Therefore, points $S,P,$ and $X$ must be collinear.

Let $G$ be the foot of the perpendicular from $D$ to $\overline{AB}.$ Note that $\overline{DG}\parallel\overline{XS},$ as shown below: $[asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q, G; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; G = foot(D,A,B); fill(D--A--G--cycle,green); fill(P--R--X--cycle,yellow); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(D,G,A),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^X--P^^D--G); draw(P--R,red+dashed); dot("$A$",A,1.5*dir(225),linewidth(4.5)); dot("$B$",B,1.5*dir(-45),linewidth(4.5)); dot("$C$",C,1.5*dir(45),linewidth(4.5)); dot("$D$",D,1.5*dir(135),linewidth(4.5)); dot("$P$",P,1.5*dir(60),linewidth(4.5)); dot("$R$",R,1.5*dir(135),linewidth(4.5)); dot("$O$",O,1.5*dir(45),linewidth(4.5)); dot("$X$",X,1.5*dir(135),linewidth(4.5)); dot("$G$",G,1.5*dir(-90),linewidth(4.5)); draw(P--X,MidArrow(0.3cm,Fill(red))); draw(G--D,MidArrow(0.3cm,Fill(red))); label("$9$",midpoint(P--R),dir(A-D),red); label("$12$",midpoint(R--X),dir(135),red); label("$15$",midpoint(X--P),dir(0),red); label("$25$",midpoint(G--D),dir(0),red); [/asy]$ As $\angle PRX = \angle AGD = 90^\circ$ and $\angle PXR = \angle ADG$ by the AA Similarity, we conclude that $\triangle PRX \sim \triangle AGD.$ The ratio of similitude is $\frac{PX}{AD} = \frac{RX}{GD}.$ We get $\frac{15}{AD} = \frac{12}{25},$ from which $AD = \frac{125}{4}.$

Finally, the perimeter of $ABCD$ is $4AD = \boxed{125}.$

~MRENTHUSIASM (inspired by awesomeming327. and WestSuburb)

Solution 3

Label the points of the rhombus to be $X$ , $Y$ , $Z$ , and $W$ and the center of the incircle to be $O$ so that $9$ , $5$ , and $16$ are the distances from point $P$ to side $ZW$ , side $WX$ , and $XY$ respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus $XYZW$ is $25$ and circle $O$ has radius $\frac{25}{2}$ .

Call the feet of the altitudes from P to side $ZW$ , side $WX$ , and side $XY$ to be $A$ , $B$ , and $C$ respectively. Additionally, call the feet of the altitudes from $O$ to side $ZW$ , side $WX$ , and side $XY$ to be $D$ , $E$ , and $F$ respectively.

Draw a line segment from $P$ to $\overline{OD}$ so that it is perpendicular to $\overline{OD}$ . Notice that this segment length is equal to $AD$ and is $\sqrt{\left(\frac{25}{2}\right)^2-\left(\frac{7}{2}\right)^2}=12$ by Pythagorean Theorem.

Similarly, perform the same operations with side $WX$ to get $BE=10$ .

By equal tangents, $WD=WE$ . Now, label the length of segment $WA=n$ and $WB=n+2$ .

Using Pythagorean Theorem again, we get

$\begin{align*} WA^2+PA^2&=WB^2+PB^2 \\ n^2+9^2&=(n+2)^2+5^2 \\ n&=13. \end{align*}$

Which also gives us $\tan{\angle{OWX}}=\frac{1}{2}$ and $OW=\frac{25\sqrt{5}}{2}$ .

Since the diagonals of the rhombus intersect at $O$ and are angle bisectors and are also perpendicular to each other, we can get that

$\begin{align*} \frac{OX}{OW}&=\tan{\angle{OWX}} \\ OX&=\frac{25\sqrt{5}}{4} \\ WX^2&=OW^2+OX^2 \\ WX&=\frac{125}{4} \\ 4WX&=\boxed{125}. \end{align*}$

~Danielzh

Solution 4

Denote by $O$ the center of $ABCD$ . We drop an altitude from $O$ to $AB$ that meets $AB$ at point $H$ . We drop altitudes from $P$ to $AB$ and $AD$ that meet $AB$ and $AD$ at $E$ and $F$ , respectively. We denote $\theta = \angle BAC$ . We denote the side length of $ABCD$ as $d$ .

Because the distances from $P$ to $BC$ and $AD$ are $16$ and $9$ , respectively, and $BC \parallel AD$ , the distance between each pair of two parallel sides of $ABCD$ is $16 + 9 = 25$ . Thus, $OH = \frac{25}{2}$ and $d \sin \theta = 25$ .

We have $\begin{align*} \angle BOH & = 90^\circ - \angle HBO \\ & = 90^\circ - \angle HBD \\ & = 90^\circ - \frac{180^\circ - \angle C}{2} \\ & = 90^\circ - \frac{180^\circ - \theta}{2} \\ & = \frac{\theta}{2} . \end{align*}$

Thus, $BH = OH \tan \angle BOH = \frac{25}{2} \tan \frac{\theta}{2}$ .

In $FAEP$ , we have $\overrightarrow{FA} + \overrightarrow{AE} + \overrightarrow{EP} + \overrightarrow{PF} = 0$ . Thus, $AF + AE e^{i \left( \pi - \theta \right)} + EP e^{i \left( \frac{3 \pi}{2} - \theta \right)} - PF i .$

Taking the imaginary part of this equation and plugging $EP = 5$ and $PF = 9$ into this equation, we get $AE = \frac{9 + 5 \cos \theta}{\sin \theta} .$

We have $\begin{align*} OP^2 & = \left( OH - EP \right)^2 + \left( AH - AE \right)^2 \\ & = \left( \frac{25}{2} - 5 \right)^2 + \left( d - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) \\ & = \left( \frac{15}{2} \right)^2 + \left( \frac{25}{\sin \theta} - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) . \hspace{1cm} (\bigstar) \end{align*}$

Because $P$ is on the incircle of $ABCD$ , $OP = \frac{25}{2}$ . Plugging this into $(\bigstar)$ , we get the following equation $20 \sin \theta - 15 \cos \theta = 7 .$

By solving this equation, we get $\sin \theta = \frac{4}{5}$ and $\cos \theta = \frac{3}{5}$ . Therefore, $d = \frac{25}{\sin \theta} = \frac{125}{4}$ .

Therefore, the perimeter of $ABCD$ is $4d = \boxed{125}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 188: / Line 188: @@
 ~Danielzh
-==Solution 3==
+==Solution 4==
 Denote by <math>O</math> the center of <math>ABCD</math>.

2023 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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