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Difference between revisions of "2019 AIME II Problems/Problem 1"

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<cmath>x=\frac{12}{5}</cmath>
 
<cmath>x=\frac{12}{5}</cmath>
  
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This means that the area is <math>A=\frac{1}{2}(9)(\frac{12}{5})=\frac{54}{5}</math>. This gets us <math>54+5=\boxed{59}.</math>
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This means that the area is <math>A=\frac{1}{2}(9)(\frac{12}{5})=\frac{54}{5}</math>. This gets us <math>54+5=\boxed{059}.</math>
 +
 
 +
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=II|before=First Problem|num-a=2}}
 
{{AIME box|year=2019|n=II|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:09, 22 March 2019

Problem

Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

[asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); draw(D--(-6,0)--A,dotted); label("$8$",(D+(-6,0))/2,dir(180)); label("$6$",(A+(-6,0))/2,dir(-90)); [/asy] - Diagram by Brendanb4321

Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so that you have a rectangle. The base $CD$ of the rectangle will be $9+6+6=21$. Now, let $E$ be the intersection of $BD$ and $AC$. This means that $\triangle ABE$ and $\triangle DCE$ are with ratio $\frac{21}{9}=\frac73$. Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $E$ to $DC$, and $x$ be the height of $\triangle ABE$. \[\frac{7}{3}=\frac{y}{x}\] \[\frac{7}{3}=\frac{8-x}{x}\] \[7x=24-3x\] \[10x=24\] \[x=\frac{12}{5}\]

This means that the area is $A=\frac{1}{2}(9)(\frac{12}{5})=\frac{54}{5}$. This gets us $54+5=\boxed{059}.$

-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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