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

(→‎Solution 1)
Line 36: Line 36:
 
</asy>
 
</asy>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
==Solution 2==
 +
Label the points of the rhombus to be <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> and the center of the incircle to be <math>O</math> so that <math>9</math>, <math>5</math>, and <math>16</math> are the distances from point <math>P</math> to side <math>ZW</math>, side <math>WX</math>, and <math>XY</math> respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus <math>XYZW</math> is <math>25</math> and circle <math>O</math> has radius <math>\frac{25}{2}</math>.
 +
 +
Call the feet of the altitudes from P to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>A</math>, <math>B</math>, and <math>C</math> respectively. Additionally, call the feet of the altitudes from <math>O</math> to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>D</math>, <math>E</math>, and <math>F</math> respectively.
 +
 +
Draw a line segment from <math>P</math> to <math>\overline{OD}</math> so that it is perpendicular to <math>\overline{OD}</math>. Notice that this segment length is equal to <math>AD</math> and is <math>\sqrt{(\frac{25}{2})^2-(\frac{7}{2})^2}=12</math> by Pythagorean Theorem.
 +
 +
Similarly, perform the same operations with side <math>WX</math> to get <math>BE=10</math>.
 +
 +
By equal tangents, <math>WD=WE</math>. Now, label the length of segment <math>WA=n</math> and <math>WB=n+2</math>.
 +
 +
Using Pythagorean Theorem again, we get
 +
 +
<cmath>
 +
\begin{align*}
 +
WA^2+PA^2&=WB^2+PB^2
 +
\\
 +
n^2+9^2&=(n+2)^2+5^2
 +
\\
 +
n&=13.
 +
\end{align*}
 +
</cmath>
 +
 +
Which also gives us <math>\tan{\angle{OWX}}=\frac{1}{2}</math> and <math>OW=\frac{25\sqrt{5}}{2}</math>.
 +
 +
Since the diagonals of the rhombus intersect at <math>O</math> and are angle bisectors and are also perpendicular to each other, we can get that
 +
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
~Danielzh
  
 
==Solution 3==
 
==Solution 3==
Line 101: Line 140:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
−
 
−
==Solution 2==
 
−
Label the points of the rhombus to be <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> and the center of the incircle to be <math>O</math> so that <math>9</math>, <math>5</math>, and <math>16</math> are the distances from point <math>P</math> to side <math>ZW</math>, side <math>WX</math>, and <math>XY</math> respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus <math>XYZW</math> is <math>25</math> and circle <math>O</math> has radius <math>\frac{25}{2}</math>.
 
−
 
−
Call the feet of the altitudes from P to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>A</math>, <math>B</math>, and <math>C</math> respectively. Additionally, call the feet of the altitudes from <math>O</math> to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>D</math>, <math>E</math>, and <math>F</math> respectively.
 
−
 
−
Draw a line segment from <math>P</math> to <math>\overline{OD}</math> so that it is perpendicular to <math>\overline{OD}</math>. Notice that this segment length is equal to <math>AD</math> and is <math>\sqrt{(\frac{25}{2})^2-(\frac{7}{2})^2}=12</math> by Pythagorean Theorem
 
−
 
−
Similarly, perform the same operations with side <math>WX</math> to get <math>BE=10</math>.
 
−
 
−
By equal tangents, <math>WD=WE</math>. Now, label the length of segment <math>WA=n</math> and <math>WB=n+2</math>
 
−
 
−
Using Pythagorean Theorem again, we get
 
−
 
−
<cmath>
 
−
\begin{align*}
 
−
WA^2+PA^2&=WB^2+PB^2
 
−
\\
 
−
n^2+9^2&=(n+2)^2+5^2
 
−
\\
 
−
n&=13
 
−
\end{align*}
 
−
</cmath>
 
−
 
−
Which also gives us <math>\tan{\angle{OWX}}=\frac{1}{2}</math> and <math>OW=\frac{25\sqrt{5}}{2}</math>
 
−
 
−
Since the diagonals of the rhombus intersect at <math>O</math> and are angle bisectors and are also perpendicular to each other, we can get that
 
−
 
−
<cmath>
 
−
\begin{align*}
 
−
\frac{OX}{OW}=\tan{\angle{OWX}}
 
−
\\
 
−
OX=\frac{25\sqrt{5}}{4}
 
−
\\
 
−
WX^2=OW^2+OX^2
 
−
\\
 
−
WX=125/4
 
−
\\
 
−
4WX=\boxed{125}
 
−
\end{align*}
 
−
</cmath>
 
−
 
−
~Danielzh
 
  
 
==See also==
 
==See also==

Revision as of 15:56, 9 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 */ 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); draw(B--T,dashed); draw(P--R^^P--S^^P--T,red); 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 2

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 $\sqrt{(\frac{25}{2})^2-(\frac{7}{2})^2}=12$ 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 3

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)

See also

2023 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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