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

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===Solution 4===
 
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(I don't know how to use Latex, so if anyone can modify my solution to add Latex, that would be helpful)
 
 
Let's call the center of the circle that segment <math>AB</math> is the diameter of, <math>O</math>. Note that <math>\triangle ODE</math> is an isosceles right triangle. Solving for side <math>OE</math>, using the Pythagorean theorem, we find it to be <math>5\sqrt{2}</math>. Calling the point where segment <math>OE</math> intersects circle <math>O</math>, the point <math>I</math>, segment <math>IE</math> would be <math>5\sqrt{2}-2</math>. Also, noting that <math>\triangle ADE</math> is a right triangle, we solve for side <math>AE</math>, using the Pythagorean Theorem, and get <math>\sqrt{74}</math>. Using Power of Point on point <math>E</math>, we can solve for <math>CE</math>. We can subtract <math>CE</math> from <math>AE</math> to find <math>AC</math> and then solve for <math>CB</math> using Pythagorean theorem once more.
 
Let's call the center of the circle that segment <math>AB</math> is the diameter of, <math>O</math>. Note that <math>\triangle ODE</math> is an isosceles right triangle. Solving for side <math>OE</math>, using the Pythagorean theorem, we find it to be <math>5\sqrt{2}</math>. Calling the point where segment <math>OE</math> intersects circle <math>O</math>, the point <math>I</math>, segment <math>IE</math> would be <math>5\sqrt{2}-2</math>. Also, noting that <math>\triangle ADE</math> is a right triangle, we solve for side <math>AE</math>, using the Pythagorean Theorem, and get <math>\sqrt{74}</math>. Using Power of Point on point <math>E</math>, we can solve for <math>CE</math>. We can subtract <math>CE</math> from <math>AE</math> to find <math>AC</math> and then solve for <math>CB</math> using Pythagorean theorem once more.
  

Revision as of 14:28, 5 September 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

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 $\boxed{\textbf{(D) } \frac{140}{37}}$.

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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