Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AIME I Problems/Problem 11"
Line 1: Line 1:
−
.
+
== Problem ==
−
==solution 1==
+
Let <math>ABCD</math> be a parallelogram with <math>\angle BAD < 90^{\circ}</math>. A circle tangent to sides <math>\overline{DA}</math>, <math>\overline{AB}</math>, and <math>\overline{BC}</math> intersects diagonal <math>\overline{AC}</math> at points <math>P</math> and <math>Q</math> with <math>AP < AQ</math>, as shown. Suppose that <math>AP = 3</math>, <math>PQ = 9</math>, and <math>QC = 16</math>. Then the area of <math>ABCD</math> can be expressed in the form <math>m\sqrt n</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
 +
 
 +
<asy>
 +
defaultpen(linewidth(0.6)+fontsize(11));
 +
size(8cm);
 +
pair A,B,C,D,P,Q;
 +
A=(0,0);
 +
label("$A$", A, SW);
 +
B=(6,15);
 +
label("$B$", B, NW);
 +
C=(30,15);
 +
label("$C$", C, NE);
 +
D=(24,0);
 +
label("$D$", D, SE);
 +
P=(5.2,2.6);
 +
label("$P$", (5.8,2.6), N);
 +
Q=(18.3,9.1);
 +
label("$Q$", (18.1,9.7), W);
 +
draw(A--B--C--D--cycle);
 +
draw(C--A);
 +
draw(Circle((10.95,7.45), 7.45));
 +
dot(A^^B^^C^^D^^P^^Q);
 +
</asy>
 +
 
 +
==Solution==
  
 
Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math>
 
Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math>
Line 7: Line 31:
  
 
Now applying LOC in <math>\triangle{ADC}</math>, getting <math>(6+x)^2+400-2(6+x)*20*\frac{2x-12}{2x+12}=(3+9+16)^2</math>, solving this equation to get <math>x=\frac{9}{2}</math>, then <math>\cos\alpha=-\frac{1}{7}</math>, <math>\sin\alpha=\frac{4\sqrt{3}}{7}</math>, the area is <math>\frac{21}{2}*\frac{49}{2}*\frac{4\sqrt{3}}{7}=147\sqrt{3}</math> leads to <math>\boxed{150}</math>
 
Now applying LOC in <math>\triangle{ADC}</math>, getting <math>(6+x)^2+400-2(6+x)*20*\frac{2x-12}{2x+12}=(3+9+16)^2</math>, solving this equation to get <math>x=\frac{9}{2}</math>, then <math>\cos\alpha=-\frac{1}{7}</math>, <math>\sin\alpha=\frac{4\sqrt{3}}{7}</math>, the area is <math>\frac{21}{2}*\frac{49}{2}*\frac{4\sqrt{3}}{7}=147\sqrt{3}</math> leads to <math>\boxed{150}</math>
 +
 +
==See Also==
 +
{{AIME box|year=2022|n=I|num-b=10|num-a=12}}
 +
{{MAA Notice}}

Revision as of 21:23, 17 February 2022

Problem

Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$. A circle tangent to sides $\overline{DA}$, $\overline{AB}$, and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$, as shown. Suppose that $AP = 3$, $PQ = 9$, and $QC = 16$. Then the area of $ABCD$ can be expressed in the form $m\sqrt n$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

[asy] defaultpen(linewidth(0.6)+fontsize(11)); size(8cm); pair A,B,C,D,P,Q; A=(0,0); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]

Solution

Let the circle tangent to $BC,AD,AB$ at $P,Q,M$ separately, denote that $\angle{ABC}=\angle{D}=\alpha$

Using POP, it is very clear that $PC=20,AQ=AM=6$, let $BM=BP=x,QD=14+x$, using LOC in $\triangle{ABP}$,$x^2+(x+6)^2-2x(x+6)\cos\alpha=36+PQ^2$, similarly, use LOC in $\triangle{DQC}$, getting that $(14+x)^2+(6+x)^2-2(6+x)(14+x)\cos\alpha=400+PQ^2$. We use the second equation to minus the first equation, getting that $28x+196-(2x+12)*14*\cos\alpha=364$, we can get $\cos\alpha=\frac{2x-12}{2x+12}$.

Now applying LOC in $\triangle{ADC}$, getting $(6+x)^2+400-2(6+x)*20*\frac{2x-12}{2x+12}=(3+9+16)^2$, solving this equation to get $x=\frac{9}{2}$, then $\cos\alpha=-\frac{1}{7}$, $\sin\alpha=\frac{4\sqrt{3}}{7}$, the area is $\frac{21}{2}*\frac{49}{2}*\frac{4\sqrt{3}}{7}=147\sqrt{3}$ leads to $\boxed{150}$

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_11&oldid=171198"