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

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(Solution 5)
 
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There is a rhombus ABCD with an incircle. A point P is chosen somewhere on the incircle,
+
==Problem==
and the distances from P to sides AB, CD, and BC, are 9, 16, and 5, respectively. Figure out the perimeter.
+
Rhombus <math>ABCD</math> has <math>\angle BAD < 90^\circ.</math> There is a point <math>P</math> on the incircle of the rhombus such that the distances from <math>P</math> to the lines <math>DA,AB,</math> and <math>BC</math> are <math>9,</math> <math>5,</math> and <math>16,</math> respectively. Find the perimeter of <math>ABCD.</math>
 +
 
 +
==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 <b>Diagram</b> section.
 +
 
 +
Let <math>O</math> be the incenter of <math>ABCD</math> for which <math>\odot O</math> is tangent to <math>\overline{DA},\overline{AB},</math> and <math>\overline{BC}</math> at <math>X,Y,</math> and <math>Z,</math> respectively. Moreover, suppose that <math>R,S,</math> and <math>T</math> are the feet of the perpendiculars from <math>P</math> to <math>\overleftrightarrow{DA},\overleftrightarrow{AB},</math> and <math>\overleftrightarrow{BC},</math> respectively, such that <math>\overline{RT}</math> intersects <math>\odot O</math> at <math>P</math> and <math>Q.</math>
 +
 
 +
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 <math>\angle RXZ = \angle TZX = 90^\circ</math> by the properties of tangents, so <math>RTZX</math> is a rectangle. It follows that the diameter of <math>\odot O</math> is <math>XZ = RT = 25.</math>
 +
 
 +
Let <math>x=PQ</math> and <math>y=RX=TZ.</math>
 +
We apply the Power of a Point Theorem to <math>R</math> and <math>T:</math>
 +
<cmath>\begin{align*}
 +
y^2 &= 9(9+x), \\
 +
y^2 &= 16(16-x).
 +
\end{align*}</cmath>
 +
We solve this system of equations to get <math>x=7</math> and <math>y=12.</math> Alternatively, we can find these results by the symmetry on rectangle <math>RTZX</math> and semicircle <math>\widehat{XPZ}.</math>
 +
 
 +
We extend <math>\overline{SP}</math> beyond <math>P</math> to intersect <math>\odot O</math> and <math>\overleftrightarrow{CD}</math> at <math>E</math> and <math>F,</math> respectively, where <math>E\neq P.</math> So, we have <math>EF=SP=5</math> and <math>PE=25-SP-EF=15.</math> On the other hand, we have <math>PX=15</math> by the Pythagorean Theorem on right <math>\triangle PRX.</math> Together, we conclude that <math>E=X.</math> Therefore, points <math>S,P,</math> and <math>X</math> must be collinear.
 +
 
 +
Let <math>G</math> be the foot of the perpendicular from <math>D</math> to <math>\overline{AB}.</math> Note that <math>\overline{DG}\parallel\overline{XP},</math> 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 <math>\angle PRX = \angle AGD = 90^\circ</math> and <math>\angle PXR = \angle ADG</math> by the AA Similarity, we conclude that <math>\triangle PRX \sim \triangle AGD.</math> The ratio of similitude is <cmath>\frac{PX}{AD} = \frac{RX}{GD}.</cmath> We get <math>\frac{15}{AD} = \frac{12}{25},</math> from which <math>AD = \frac{125}{4}.</math>
 +
 
 +
Finally, the perimeter of <math>ABCD</math> is <math>4AD = \boxed{125}.</math>
 +
 
 +
~MRENTHUSIASM (inspired by awesomeming327. and WestSuburb)
 +
 
 +
==Solution 2==
 +
 
 +
This solution refers to the <b>Diagram</b> section.
 +
 
 +
Define points <math>O,R,S,</math> and <math>T</math> as Solution 1 does. Moreover, let
 +
<math>H</math> be the foot of the perpendicular from <math>P</math> to <math>\overleftrightarrow{CD},</math>
 +
<math>M</math> be the foot of the perpendicular from <math>O</math> to <math>\overleftrightarrow{HS},</math> and
 +
<math>N</math> be the foot of the perpendicular from <math>O</math> to <math>\overleftrightarrow{RT}.</math>
 +
 
 +
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, H, M, N;
 +
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);
 +
H = foot(S,C,D);
 +
M = foot(O,S,H);
 +
N = foot(O,R,T);
 +
 
 +
fill(O--M--P--cycle,yellow);
 +
fill(O--N--P--cycle,green);
 +
markscalefactor=0.15;
 +
draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red);
 +
draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P);
 +
draw(P--R^^P--S^^P--T^^P--H,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(90),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("$H$",H,1.5*dir(90),linewidth(4.5));
 +
dot("$M$",M,1.5*dir(180),linewidth(4.5));
 +
dot("$N$",N,1.5*dir(15),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 the diameter of <math>\odot O</math> is <math>HS=RT=25,</math> so <math>OP=\frac{25}{2}.</math> It follows that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>In right <math>\triangle OMP,</math> we have <math>MP=\frac{HS}{2}-PS=\frac{15}{2}</math> by symmetry, from which <math>OM=10</math> by the Pythagorean Theorem.</li><p>
 +
  <li>In right <math>\triangle ONP,</math> we have <math>NP=\frac{RT}{2}-RP=\frac{7}{2}</math> by symmetry, from which <math>ON=12</math> by the Pythagorean Theorem.</li><p>
 +
</ol>
 +
Since <math>\overline{MO}\parallel\overline{AB}</math> and <math>\overline{ON}\parallel\overline{DA},</math> we conclude that <math>\angle A = \angle MON.</math> We apply the Sine of a Sum Formula:
 +
<cmath>\begin{align*}
 +
\sin\angle A &= \sin\angle MON \\
 +
&= \sin(\angle MOP + \angle PON) \\
 +
&= \sin\angle MOP \cos\angle PON + \cos\angle MOP \sin\angle PON \\
 +
&= \frac{3}{5}\cdot\frac{24}{25} + \frac{4}{5}\cdot\frac{7}{25} \\
 +
&= \frac{4}{5}.
 +
\end{align*}</cmath>
 +
Note that <cmath>\sin\angle A = \frac{HS}{DA},</cmath> from which <math>\frac{4}{5} = \frac{25}{DA}.</math> We solve this equation to get <math>DA=\frac{125}{4}.</math>
 +
 
 +
Finally, the perimeter of <math>ABCD</math> is <math>4DA = \boxed{125}.</math>
 +
 
 +
~MRENTHUSIASM (credit given to TheAMCHub)
 +
 
 +
==Solution 3==
 +
 
 +
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 <math>P</math> 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{\left(\frac{25}{2}\right)^2-\left(\frac{7}{2}\right)^2}=12</math> by Pythagorean Theorem.
 +
 
 +
Similarly, perform the same operations with perpendicular from <math>P</math> to <math>\overline{OE}</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>
 +
 
 +
~[[Daniel Zhou's Profile|Danielzh]]
 +
 
 +
==Solution 4==
 +
 
 +
Denote by <math>O</math> the center of <math>ABCD</math>.
 +
We drop an altitude from <math>O</math> to <math>AB</math> that meets <math>AB</math> at point <math>H</math>.
 +
We drop altitudes from <math>P</math> to <math>AB</math> and <math>AD</math> that meet <math>AB</math> and <math>AD</math> at <math>E</math> and <math>F</math>, respectively.
 +
We denote <math>\theta = \angle BAC</math>.
 +
We denote the side length of <math>ABCD</math> as <math>d</math>.
 +
 
 +
Because the distances from <math>P</math> to <math>BC</math> and <math>AD</math> are <math>16</math> and <math>9</math>, respectively, and <math>BC \parallel AD</math>, the distance between each pair of two parallel sides of <math>ABCD</math> is <math>16 + 9 = 25</math>.
 +
Thus, <math>OH = \frac{25}{2}</math> and <math>d \sin \theta = 25</math>.
 +
 
 +
We have
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 
 +
Thus, <math>BH = OH \tan \angle BOH = \frac{25}{2} \tan \frac{\theta}{2}</math>.
 +
 
 +
In <math>FAEP</math>, we have <math>\overrightarrow{FA} + \overrightarrow{AE} + \overrightarrow{EP} + \overrightarrow{PF} = 0</math>.
 +
Thus,
 +
<cmath>
 +
\[
 +
AF + AE e^{i \left( \pi - \theta \right)} + EP e^{i \left( \frac{3 \pi}{2} - \theta \right)}
 +
- PF i .
 +
\]
 +
</cmath>
 +
 
 +
Taking the imaginary part of this equation and plugging <math>EP = 5</math> and <math>PF = 9</math> into this equation, we get
 +
<cmath>
 +
\[
 +
AE = \frac{9 + 5 \cos \theta}{\sin \theta} .
 +
\]
 +
</cmath>
 +
 
 +
We have
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 
 +
Because <math>P</math> is on the incircle of <math>ABCD</math>, <math>OP = \frac{25}{2}</math>. Plugging this into <math>(\bigstar)</math>, we get the following equation
 +
<cmath>
 +
\[
 +
20 \sin \theta - 15 \cos \theta = 7 .
 +
\]
 +
</cmath>
 +
 
 +
By solving this equation, we get <math>\sin \theta = \frac{4}{5}</math> and <math>\cos \theta = \frac{3}{5}</math>.
 +
Therefore, <math>d = \frac{25}{\sin \theta} = \frac{125}{4}</math>.
 +
 
 +
Therefore, the perimeter of <math>ABCD</math> is <math>4d = \boxed{125}</math>.
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Solution 5==
 +
 
 +
The center of the incircle is <math>O.</math> Denote the points in which the incircle meets <math>\overline{AB},</math> <math>\overline{BC},</math> <math>\overline{CD},</math> and <math>\overline{DA}</math> as <math>W,</math> <math>X,</math> <math>Y,</math> and <math>Z,</math> respectively. Next, also denote the base of the perpendicular from <math>P</math> to <math>\overline{AB},</math> <math>\overline{AD},</math> <math>\overline{OW},</math> and <math>\overline{OZ}</math> as <math>M,</math> <math>N,</math> <math>S,</math> and <math>T,</math> respectively.
 +
 
 +
We can easily see that the radius of the circle is <math>\frac{25}{2}.</math> Using this and Pythagorus on right <math>\triangle OSP</math> and <math>\triangle OTP,</math> we find that <math>MW = PS = 10</math> and <math>NZ = PT = 12.</math>
 +
 
 +
Since <math>AW = AZ</math> by properties of circle tangents, we can deduce by the above information that <math>AM = AN+2.</math> Doing Pythagorus on right <math>\triangle AMP</math> and <math>\triangle ANP</math> we find that <math>a^2 = b^2 + 56</math> (because <math>a^2+25=b^2+81.</math>) From solving the <math>2</math> just derived equations, we find that <math>AM=15</math> and <math>AN=13.</math>
 +
 
 +
Next, we use Pythagorus on right <math>\triangle AOB</math> (we can see it's right because of properties of rhombuses.) We get <cmath>AB^2 = AO^2 + BO^2.</cmath> We know <math>AB = AW + WB = 25 + WB.</math> By Pythagorus on <math>\triangle AWO</math> and <math>\triangle BWO,</math> we also know <math>AO^2 = 25^2+\left(\frac{25}{2}\right)^2</math> and <math>BO^2=WB^2+\left(\frac{25}{2}\right)^2.</math> Substituting these in, we have <cmath>25^2 + 50WB + WB^2 = 25^2+\left(\frac{25}{2}\right)^2+\left(\frac{25}{2}\right)^2+WB^2.</cmath> Solving for <math>WB,</math> we get <math>WB = \frac{25}{4}.</math> Now we find that each side of the rhombus <math>=AB=25+\frac{25}{4}=\frac{125}{4}.</math> The perimeter of the rhombus would be that times <math>4.</math> Our final answer is <cmath>\frac{125}{4}\cdot4=\boxed{125}.</cmath>
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~s214425
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==Solution 6 ==
 +
[[File:2023 AIME I 8.png|450px|right]]
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Notation is shown on diagram, <math>RT \perp AD, FG \perp AB, E = AD \cap \omega, E' = FG \cap AD.</math>
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<math>RT = 9 + 16 = 25 = FG</math> as hights of rhombus.
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<cmath>RP = QT = 9, PQ = 16 - 9 = 7, GE' = PF = 5,</cmath>
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<cmath>PE' = 25 - 5 - 5 = 15, RE = \sqrt{RP \cdot RQ} = \sqrt{9 \cdot 16} = 12.</cmath>
 +
<cmath>PE =  \sqrt{RP^2 + RE^2} = 15 \implies E = E'.</cmath>
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<cmath>\sin \alpha = \frac {RE}{PE} =  \frac {GF}{AD} \implies AD = \frac {15 \cdot 25}{12} = \frac {125}{4}.</cmath>
 +
The perimeter of <math>ABCD </math> is <math>\frac{125}{4}\cdot4=\boxed{125}.</math>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Video Solution==
 +
https://youtu.be/AYH6zdJqZLM
 +
 
 +
~MathProblemSolvingSkills.com
  
 
==See also==
 
==See also==

Latest revision as of 13:54, 3 July 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 $\overline{DG}\parallel\overline{XP},$ 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 2

This solution refers to the Diagram section.

Define points $O,R,S,$ and $T$ as Solution 1 does. Moreover, let $H$ be the foot of the perpendicular from $P$ to $\overleftrightarrow{CD},$ $M$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{HS},$ and $N$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{RT}.$

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, H, M, N; 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); H = foot(S,C,D); M = foot(O,S,H); N = foot(O,R,T);  fill(O--M--P--cycle,yellow); fill(O--N--P--cycle,green); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P); draw(P--R^^P--S^^P--T^^P--H,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(90),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("$H$",H,1.5*dir(90),linewidth(4.5)); dot("$M$",M,1.5*dir(180),linewidth(4.5)); dot("$N$",N,1.5*dir(15),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 the diameter of $\odot O$ is $HS=RT=25,$ so $OP=\frac{25}{2}.$ It follows that:

  1. In right $\triangle OMP,$ we have $MP=\frac{HS}{2}-PS=\frac{15}{2}$ by symmetry, from which $OM=10$ by the Pythagorean Theorem.
  2. In right $\triangle ONP,$ we have $NP=\frac{RT}{2}-RP=\frac{7}{2}$ by symmetry, from which $ON=12$ by the Pythagorean Theorem.

Since $\overline{MO}\parallel\overline{AB}$ and $\overline{ON}\parallel\overline{DA},$ we conclude that $\angle A = \angle MON.$ We apply the Sine of a Sum Formula: \begin{align*} \sin\angle A &= \sin\angle MON \\ &= \sin(\angle MOP + \angle PON) \\ &= \sin\angle MOP \cos\angle PON + \cos\angle MOP \sin\angle PON \\ &= \frac{3}{5}\cdot\frac{24}{25} + \frac{4}{5}\cdot\frac{7}{25} \\ &= \frac{4}{5}. \end{align*} Note that \[\sin\angle A = \frac{HS}{DA},\] from which $\frac{4}{5} = \frac{25}{DA}.$ We solve this equation to get $DA=\frac{125}{4}.$

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

~MRENTHUSIASM (credit given to TheAMCHub)

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 perpendicular from $P$ to $\overline{OE}$ 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)

Solution 5

The center of the incircle is $O.$ Denote the points in which the incircle meets $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ and $\overline{DA}$ as $W,$ $X,$ $Y,$ and $Z,$ respectively. Next, also denote the base of the perpendicular from $P$ to $\overline{AB},$ $\overline{AD},$ $\overline{OW},$ and $\overline{OZ}$ as $M,$ $N,$ $S,$ and $T,$ respectively.

We can easily see that the radius of the circle is $\frac{25}{2}.$ Using this and Pythagorus on right $\triangle OSP$ and $\triangle OTP,$ we find that $MW = PS = 10$ and $NZ = PT = 12.$

Since $AW = AZ$ by properties of circle tangents, we can deduce by the above information that $AM = AN+2.$ Doing Pythagorus on right $\triangle AMP$ and $\triangle ANP$ we find that $a^2 = b^2 + 56$ (because $a^2+25=b^2+81.$) From solving the $2$ just derived equations, we find that $AM=15$ and $AN=13.$

Next, we use Pythagorus on right $\triangle AOB$ (we can see it's right because of properties of rhombuses.) We get \[AB^2 = AO^2 + BO^2.\] We know $AB = AW + WB = 25 + WB.$ By Pythagorus on $\triangle AWO$ and $\triangle BWO,$ we also know $AO^2 = 25^2+\left(\frac{25}{2}\right)^2$ and $BO^2=WB^2+\left(\frac{25}{2}\right)^2.$ Substituting these in, we have \[25^2 + 50WB + WB^2 = 25^2+\left(\frac{25}{2}\right)^2+\left(\frac{25}{2}\right)^2+WB^2.\] Solving for $WB,$ we get $WB = \frac{25}{4}.$ Now we find that each side of the rhombus $=AB=25+\frac{25}{4}=\frac{125}{4}.$ The perimeter of the rhombus would be that times $4.$ Our final answer is \[\frac{125}{4}\cdot4=\boxed{125}.\]

~s214425

Solution 6

2023 AIME I 8.png

Notation is shown on diagram, $RT \perp AD, FG \perp AB, E = AD \cap \omega, E' = FG \cap AD.$ $RT = 9 + 16 = 25 = FG$ as hights of rhombus. \[RP = QT = 9, PQ = 16 - 9 = 7, GE' = PF = 5,\] \[PE' = 25 - 5 - 5 = 15, RE = \sqrt{RP \cdot RQ} = \sqrt{9 \cdot 16} = 12.\] \[PE =  \sqrt{RP^2 + RE^2} = 15 \implies E = E'.\] \[\sin \alpha = \frac {RE}{PE} =  \frac {GF}{AD} \implies AD = \frac {15 \cdot 25}{12} = \frac {125}{4}.\] The perimeter of $ABCD$ is $\frac{125}{4}\cdot4=\boxed{125}.$

vladimir.shelomovskii@gmail.com, vvsss

Video Solution

https://youtu.be/AYH6zdJqZLM

~MathProblemSolvingSkills.com

See also

2023 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png