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

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<math>\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}</math>
 
<math>\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}</math>
  
−
==Solutions==
+
==Solution 1==
 +
Notice that <math>ADE</math> and <math>ABC</math> are right triangles. Then <math>AE = \sqrt{7^2+5^2} = \sqrt{74}</math>. <math>\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}</math>, so <math>BC = \frac{20}{\sqrt{74}}</math>. We also find that <math>AC = \frac{28}{\sqrt{74}}</math> (You can also use power of point ~MATHWIZARD2010), and thus the area of <math>ABC</math> is <math>\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}</math>.
 +
 
 
<center><asy>
 
<center><asy>
 
size(10cm);
 
size(10cm);
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label("$5$",(D+E)/2,NE);
 
label("$5$",(D+E)/2,NE);
 
</asy></center>
 
</asy></center>
−
===Solution 1===
 
−
Notice that <math>ADE</math> and <math>ABC</math> are right triangles. Then <math>AE = \sqrt{7^2+5^2} = \sqrt{74}</math>. <math>\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}</math>, so <math>BC = \frac{20}{\sqrt{74}}</math>. We also find that <math>AC = \frac{28}{\sqrt{74}}</math>, and thus the area of <math>ABC</math> is <math>\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}</math>.
 
  
−
===Solution 2===
+
==Solution 2 (Similar Triangles) ==
−
We note that <math>\triangle ACB ~ \triangle ADE</math> by <math>AA</math> similarity. Also, since the area of <math>\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2</math> and <math>AE = \sqrt{74}</math>, <math>\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = \left(\frac{4}{\sqrt{74}}\right)^2</math>, so the area of <math>\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}</math>.
+
We note that <math>\triangle ACB \sim \triangle ADE</math> by <math>AA</math> similarity. Also, since the area of <math>\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2</math> and <math>AE = \sqrt{74}</math>, <math>\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = \left(\frac{4}{\sqrt{74}}\right)^2</math>, so the area of <math>\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}</math>.
  
−
===Solution 3===
+
==Solution 3==
−
As stated before, note that <math>\triangle ACB ~ \triangle ADE</math>. By similarity, we note that <math>\frac{\overline{AC}}{\overline{BC}}</math> is equivalent to <math>\frac{7}{5}</math>. We set <math>\overline{AC}</math> to <math>7x</math> and <math>\overline{BC}</math> to <math>5x</math>. By the Pythagorean Theorem, <math>(7x)^2+(5x)^2 = 4^2</math>. Combining, <math>49x^2+25x^2=16</math>. We can add and divide to get <math>x^2=\frac{8}{37}</math>. We square root and rearrange to get <math>x=\frac{2\sqrt{74}}{37}</math>. We know that the legs of the triangle are <math>7x</math> and <math>5x</math>. Mulitplying <math>x</math> by <math>7</math> and <math>5</math> eventually gives us <math>\frac {14\sqrt{74}}{37}</math> <math>\frac {10\sqrt{74}}{37}</math>. We divide this by 2, since <math>\frac{1}{2}bh</math> is the formula for a triangle. This gives us <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>.
+
As stated before, note that <math>\triangle ACB</math> is similar to <math>\triangle ADE</math>. By similarity, we note that <math>\frac{\overline{AC}}{\overline{BC}}</math> is equivalent to <math>\frac{7}{5}</math>. We set <math>\overline{AC}</math> to <math>7x</math> and <math>\overline{BC}</math> to <math>5x</math>. By the Pythagorean Theorem, <math>(7x)^2+(5x)^2 = 4^2</math>. Combining, <math>49x^2+25x^2=16</math>. We can add and divide to get <math>x^2=\frac{8}{37}</math>. We square root and rearrange to get <math>x=\frac{2\sqrt{74}}{37}</math>. We know that the legs of the triangle are <math>7x</math> and <math>5x</math>. Multiplying <math>x</math> by <math>7</math> and <math>5</math> eventually gives us <math>\frac {14\sqrt{74}}{37}</math> and <math>\frac {10\sqrt{74}}{37}</math>. We divide this by <math>2</math>, since <math>\frac{1}{2}bh</math> is the formula for a triangle. This gives us <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>.
  
−
===Solution 4===
+
==Solution 4==
 
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.
  
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Note that <math>\triangle ABC</math> is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases <math>AC</math> and <math>BC</math>, we get the area of triangle <math>ABC</math> to be <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>.
 
Note that <math>\triangle ABC</math> is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases <math>AC</math> and <math>BC</math>, we get the area of triangle <math>ABC</math> to be <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>.
 +
 +
== Solution 5 (Coordinate Geo)==
 +
 +
Drawing the picture, we realize that the equation for the line from A to E is <math>y=\frac{5x}{7}</math>, and the equation for the circle is <math>(x-2)^2+y^2=4</math>
 +
plugging in <math>\frac{5x}{7}</math> for y we get <math>x(74x-196)=0</math> so <math>x=\frac{98}{37}</math>, that means <math>y = \frac{98}{37} \cdot \frac{5}{7} = \frac{70}{37}</math>
 +
 +
the height is <math>\frac{70}{37}</math> and the base is <math>4</math>, so the area is <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>
 +
 +
-harsha12345
 +
 +
== Solution 6 (Coordinate bashing) ==
 +
We draw out the diagram, and let the center of the circle be the origin. <math>A</math> would then be <math>(0,-2)</math>, and <math>B</math> would be <math>(2,0)</math>. We find that the equation of the line <math>AE</math> is <math>\frac{5}{7}x + \frac{10}{7}</math>. The equation of the circle is <math>x^2+y^2=4</math>. We use substitution and bashing with the quadratic formula to get <math>x = \frac{24}{37}</math>. From this, we get <math>y = \frac{70}{37}</math> and get that the area is <math>\frac{70}{37}\cdot 4/2 = \boxed{\textbf{(D) } \frac{140}{37}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}}
 
{{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:16, 13 May 2024

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 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 $AC = \frac{28}{\sqrt{74}}$ (You can also use power of point ~MATHWIZARD2010), and thus the area of $ABC$ is $\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}$.

[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 2 (Similar Triangles)

We note that $\triangle ACB \sim \triangle ADE$ 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 $\triangle ACB$ is similar to $\triangle ADE$. 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 $5x$. Multiplying $x$ by $7$ and $5$ eventually gives us $\frac {14\sqrt{74}}{37}$ and $\frac {10\sqrt{74}}{37}$. 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 $O$ + $IE$)$(IE)$ $\rightarrow$ ${\sqrt{74}}(CE)$ = $(5\sqrt{2}+2)(5\sqrt{2}-2)$ $\Rightarrow$ $CE$ = $\frac{23\sqrt{74}}{37}$

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

Now to solve for $CB$:

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

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

Solution 5 (Coordinate Geo)

Drawing the picture, we realize that the equation for the line from A to E is $y=\frac{5x}{7}$, and the equation for the circle is $(x-2)^2+y^2=4$ plugging in $\frac{5x}{7}$ for y we get $x(74x-196)=0$ so $x=\frac{98}{37}$, that means $y = \frac{98}{37} \cdot \frac{5}{7} = \frac{70}{37}$

the height is $\frac{70}{37}$ and the base is $4$, so the area is $\boxed{\textbf{(D) } \frac{140}{37}}$

-harsha12345

Solution 6 (Coordinate bashing)

We draw out the diagram, and let the center of the circle be the origin. $A$ would then be $(0,-2)$, and $B$ would be $(2,0)$. We find that the equation of the line $AE$ is $\frac{5}{7}x + \frac{10}{7}$. The equation of the circle is $x^2+y^2=4$. We use substitution and bashing with the quadratic formula to get $x = \frac{24}{37}$. From this, we get $y = \frac{70}{37}$ and get that the area is $\frac{70}{37}\cdot 4/2 = \boxed{\textbf{(D) } \frac{140}{37}}$.

See Also

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

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