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

## 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 $A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}$. 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

## Solution 2

Using the diagram in Solution 1, let $E$ be the intersection of $BD$ and $AC$. We can see that angle $C$ is in both $\triangle BCE$ and $\triangle ABC$. Since $\triangle BCE$ and $\triangle ADE$ are congruent by AAS, we can then state $AE=BE$ and $DE=CE$. It follows that $BE=AE$ and $CE=17-BE$. We can now state that the area of $\triangle ABE$ is the area of $\triangle ABC-$ the area of $\triangle BCE$. Using Heron's formula, we compute the area of $\triangle ABC=36$. Using the Law of Cosines on angle $C$, we obtain $$9^2=17^2+10^2-2(17)(10)cosC$$ $$-308=-340cosC$$ $$cosC=\frac{308}{340}$$ (For convenience, we're not going to simplify.)

Applying the Law of Cosines on $\triangle BCE$ yields $$BE^2=10^2+(17-BE)^2-2(10)(17-BE)cosC$$ $$BE^2=389-34BE+BE^2-20(17-BE)(\frac{308}{340})$$ $$0=389-34BE-(340-20BE)(\frac{308}{340})$$ $$0=389-34BE+\frac{308BE}{17}$$ $$0=81-\frac{270BE}{17}$$ $$81=\frac{270BE}{17}$$ $$BE=\frac{51}{10}$$ This means $CE=17-BE=17-\frac{51}{10}=\frac{119}{10}$. Next, apply Heron's formula to get the area of $\triangle BCE$, which equals $\frac{126}{5}$ after simplifying. Subtracting the area of $\triangle BCE$ from the area of $\triangle ABC$ yields the area of $\triangle ABE$, which is $\frac{54}{5}$, giving us our answer, which is $54+5=\boxed{059}.$ -Solution by flobszemathguy

## Solution 3 (Very quick) $[asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); draw(A--B--C--cycle); draw(B--D--A); label("A",A,dir(-120)); label("B",B,dir(-60)); label("C",C,dir(60)); label("D",D,dir(120)); label("9",(A+B)/2,dir(-90)); label("10",(D+A)/2,dir(-150)); label("10",(C+B)/2,dir(-30)); label("17",(D+B)/2,dir(60)); label("17",(A+C)/2,dir(120)); draw(D--(-6,0)--A,dotted); label("8",(D+(-6,0))/2,dir(180)); label("6",(A+(-6,0))/2,dir(-90)); draw((4.5,0)--(4.5,2.4),dotted); label("h", (4.5,1.2), dir(180)); label("4.5", (6,0), dir(90)); [/asy]$ - Diagram by Brendanb4321 extended by Duoquinquagintillion

Begin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$. Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$, so the other leg of the new triangle formed has length $4.5$. Notice we have formed similar triangles, and we can solve for $h$. $$\frac{h}{4.5} = \frac{8}{15}$$ $$h = \frac{36}{15} = \frac{12}{5}$$

So $\triangle ABE$ has area $$\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}$$ And $54+5=\boxed{059}.$ - Solution by Duoquinquagintillion

## Solution 4

Let $a = \angle{CAB}$. By Law of Cosines, $$\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}$$ $$\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}$$ $$\tan a = \frac{8}{15}$$ $$A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}$$ And $54+5=\boxed{059}.$

- by Mathdummy

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