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

m (Solution)
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Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is the hypotenuse and connect the points <math>CD</math> so  
 
Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is the hypotenuse and connect the points <math>CD</math> so  
that you have a rectangle. (We know that <math>\triangle ADE</math> is a <math>6-8-10</math>, since <math>\triangle DEB</math> is an <math>8-15-17</math>.) The base <math>CD</math> of the rectangle will be <math>9+6+6=21</math>. Now, let <math>E</math> be the intersection of <math>BD</math> and <math>AC</math>. This means that <math>\triangle ABE</math> and <math>\triangle DCE</math> are with ratio <math>\frac{21}{9}=\frac73</math>. Set up a proportion, knowing that the two heights add up to 8. We will let <math>y</math> be the height from <math>E</math> to <math>DC</math>, and <math>x</math> be the height of <math>\triangle ABE</math>.
+
that you have a rectangle. (We know that <math>\triangle ADE</math> is a <math>6-8-10</math>, since <math>\triangle DEB</math> is an <math>8-15-17</math>.) The base <math>CD</math> of the rectangle will be <math>9+6+6=21</math>. Now, let <math>O</math> be the intersection of <math>BD</math> and <math>AC</math>. This means that <math>\triangle ABO</math> and <math>\triangle DCO</math> are with ratio <math>\frac{21}{9}=\frac73</math>. Set up a proportion, knowing that the two heights add up to 8. We will let <math>y</math> be the height from <math>O</math> to <math>DC</math>, and <math>x</math> be the height of <math>\triangle ABO</math>.
 
<cmath>\frac{7}{3}=\frac{y}{x}</cmath>
 
<cmath>\frac{7}{3}=\frac{y}{x}</cmath>
 
<cmath>\frac{7}{3}=\frac{8-x}{x}</cmath>
 
<cmath>\frac{7}{3}=\frac{8-x}{x}</cmath>

Latest revision as of 18:08, 26 December 2020

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); pair E = (-6,0); 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("$E$",E,dir(-135)); 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--E--A,dotted); label("$8$",(D+E)/2,dir(180)); label("$6$",(A+E)/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. (We know that $\triangle ADE$ is a $6-8-10$, since $\triangle DEB$ is an $8-15-17$.) The base $CD$ of the rectangle will be $9+6+6=21$. Now, let $O$ be the intersection of $BD$ and $AC$. This means that $\triangle ABO$ and $\triangle DCO$ 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 $O$ to $DC$, and $x$ be the height of $\triangle ABO$. \[\frac{7}{3}=\frac{y}{x}\] \[\frac{7}{3}=\frac{8-x}{x}\] \[7x=24-3x\] \[10x=24\] \[x=\frac{12}{5}\]

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

Solution 5

Because $AD = BC$ and $\angle BAD = \angle ABC$, quadrilateral $ABCD$ is cyclic. So, Ptolemy's theorem tells us that \[AB \cdot CD + BC \cdot AD = AC \cdot BD \implies 9 \cdot CD + 10^2 = 17^2 \implies CD = 21.\]

From here, there are many ways to finish which have been listed above. If we let $AB \cap CD = P$, then \[\triangle APB \sim \triangle CPD \implies \frac{AP}{AB} = \frac{CP}{CD} \implies \frac{AP}{9} = \frac{17-AP}{21} \implies AP = 5.1.\]

Using Heron's formula on $\triangle ABP$, we see that \[[ABC] = \sqrt{9.6(9.6-5.1)(9.6-5.1)(9.6-9)} = 10.8 = \frac{54}{5}.\]

Thus, our answer is $059$. ~a.y.711

See Also

2019 AIME II (ProblemsAnswer KeyResources)
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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