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

Revision as of 09:30, 28 September 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 $A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}$ . 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 Consine, $\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}$ $\sin a = \sqrt{1-\cos a} = \frac{8}{17}$ $\tan a = \frac{8}{15}$ $A = \frac{1}{2}* \frac{9}{2}*\frac{9}{2}\tan a = \frac{54}{5}$ And $54+5=\boxed{059}.$

- by Mathdummy

@@ Line 99: / Line 99: @@
 And <math>54+5=\boxed{059}.</math>
 - Solution by Duoquinquagintillion
+== Solution 4 ==
+Let <math>a = \angle{CAB}</math>. By Law of Consine,
+<cmath>\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}</cmath>
+<cmath>\sin a = \sqrt{1-\cos a} = \frac{8}{17}</cmath>
+<cmath>\tan a = \frac{8}{15}</cmath>
+<cmath>A = \frac{1}{2}* \frac{9}{2}*\frac{9}{2}\tan a = \frac{54}{5}</cmath>
+And <math>54+5=\boxed{059}.</math>
+- by Mathdummy
 ==See Also==
 {{AIME box|year=2019|n=II|before=First Problem|num-a=2}}
 {{MAA Notice}}

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