Difference between revisions of "2017 AMC 10B Problems/Problem 22"

Revision as of 18:47, 12 December 2017

Problem

The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$ . Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$ . Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E$ . What is the area of $\triangle ABC$ ?

$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$

Solution

$[asy] size(10cm); pair A, B, C, D, E, O; A = (-2,0); B = (2,0); C = (2*cos(1.24),2*sin(1.24)); D = (5,0); E = (5,5); O = (A+B)/2; dot(A); dot(B); dot(C); dot(D); dot(E); dot(O); draw(Circle((A+B)/2,2)); draw(A--D--E--C--A); draw(C--B); draw(rightanglemark(A,C,B,5)); draw(rightanglemark(A,D,E,5)); label("$A$",A,W); label("$B$",B,SE); label("$D$",D,SE); label("$E$",E,NE); label("$C$",C,N); label("$2$",(O+B)/2,S); label("$3$",(B+D)/2,S); label("$5$",(D+E)/2,NE); [/asy]$

Solution 1

Notice that $ADE$ and $ABC$ are right triangles. Then $AE = \sqrt{7^2+5^2} = \sqrt{74}$ . $\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}$ , so $BC = \frac{20}{\sqrt{74}}$ . We also find that $AC = \frac{28}{\sqrt{74}}$ , and thus the area of $ABC$ is $\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}$ .

Solution 2

We note that $\triangle ACB ~ \triangle ADE$ by $AA$ similarity. Also, since the area of $\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2$ and $AE = \sqrt{74}$ , $\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = \left(\frac{4}{\sqrt{74}}\right)^2$ , so the area of $\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}$ .

Solution 3

As stated before, note that $\triangle ACB ~ \triangle ADE$ . By similarity, we note that $\frac{\overline{AC}}{\overline{BC}}$ is equivalent to $\frac{7}{5}$ . We set $\overline{AC}$ to $7x$ and $\overline{BC}$ to $5x$ . By the Pythagorean Theorem, $(7x)^2+(5x)^2$ = 4^2. Combining, $49x^2+25x^2=16$ . We can add and divide to get $x^2=\frac{8}{37}$ . We square root and rearrange to get $x=\frac{2\sqrt{74}}{37}$ . We know that the legs of the triangle are $7x$ and $5x$ . Mulitplying $x$ by 7 and 5 eventually gives us $\frac{14\sqrt{74}}{37}$ x $\frac{10\sqrt{74}}{37}$ . We divide this by 2, since $\frac{1}{2}bh$ is the formula for a triangle. This gives us $\boxed{\textbf{(D) } \frac{140}{37}}$ .

Solution 4

Let's call the center of the circle that segment $AB$ is the diameter of, $O$ . Note that $\triangle ODE$ is an isosceles right triangle. Solving for side $OE$ , using the Pythagorean theorem, we find it to be $5\sqrt{2}$ . Calling the point where segment $OE$ intersects circle $O$ , the point $I$ , segment $IE$ would be $5\sqrt{2}-2$ . Also, noting that $\triangle ADE$ is a right triangle, we solve for side $AE$ , using the Pythagorean Theorem, and get $\sqrt{74}$ . Using Power of Point on point $E$ , we can solve for $CE$ . We can subtract $CE$ from $AE$ to find $AC$ and then solve for $CB$ using Pythagorean theorem once more.

$(AE)(CE)$ = (Diameter of circle $O$ + $IE$ ) $(IE)$ $->$ ${\sqrt{74}}(CE)$ = $(5\sqrt{2}+2)(5\sqrt{2}-2)$ $->$ $CE$ = $\frac{23\sqrt{74}}{37}$

$AC = AE - CE$ $->$ $AC$ = ${\sqrt74}$ - $\frac{23\sqrt{74}}{37}$ $->$ $AC$ = $\frac{14\sqrt{74}}{37}$

Now to solve for $CB$ :

$AB^2$ - $AC^2$ = $CB^2$ $->$ $4^2$ + $\frac{14\sqrt{74}}{37}^2$ = $CB^2$ $->$ $CB$ = $\frac{10\sqrt{74}}{37}$

Note that $\triangle ABC$ is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases $AC$ and $BC$ , we get the area of triangle $ABC$ to be $\boxed{\textbf{(D) } \frac{140}{37}}$ .

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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