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Difference between revisions of "2009 AMC 10B Problems/Problem 18"
(→‎Solution)
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== Solution ==
 
== Solution ==
  
 +
== Solution 1 ==
 
<asy>
 
<asy>
 
unitsize(0.75cm);
 
unitsize(0.75cm);
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And as the area of triangle <math>ABC</math> is <math>\frac{6\cdot 8}2 = 24</math>, the area of triangle <math>AME</math> is <math>24\cdot \frac{25}{64} = \boxed{ \frac{75}8 }</math>.
 
And as the area of triangle <math>ABC</math> is <math>\frac{6\cdot 8}2 = 24</math>, the area of triangle <math>AME</math> is <math>24\cdot \frac{25}{64} = \boxed{ \frac{75}8 }</math>.
 +
 +
== Solution 2 (Only Pythagorean Theorem) ==
 +
<asy>
 +
unitsize(0.75cm);
 +
defaultpen(0.8);
 +
pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;
 +
path ortho = shift(M)*rotate(-90)*(A--C);
 +
pair Ep = intersectionpoint(ortho, A--B);
 +
draw( A--B--C--D--cycle );
 +
draw( A--C );
 +
draw( M--Ep );
 +
filldraw( A--M--Ep--cycle, lightgray, black );
 +
draw( rightanglemark(A,M,Ep) );
 +
draw( C--Ep );
 +
label("$A$",A,SW);
 +
label("$B$",B,SE);
 +
label("$C$",C,NE);
 +
label("$D$",D,NW);
 +
label("$E$",Ep,S);
 +
label("$M$",M,NW);
 +
</asy>
 +
 +
Draw <math>EC</math> as shown from the diagram. Since <math>AC</math> is of length <math>10</math>, we have that <math>AM</math> is of length <math>5</math>, because of the midpoint <math>M</math>. Through the Pythagorean theorem, we know that <math>AE^2 = AM^2 + ME^2 \implies 25 + ME^2</math>, which means <math>AE = \sqrt{25 + ME^2}</math>. Define <math>ME</math> to be <math>x</math> for the sake of clarity. We know that <math>EB = 8 - \sqrt{25 + x^2}</math>. From here, we know that <math>CE^2 = CB^2 + BE^2 = ME^2 + MC^2</math>. From here, we can write the expression <math>6^2 + (8 - \sqrt{25 + x^2})^2 = 5^2 + x^2 \implies 36 + (64 - 16\sqrt{25 + x^2} + 25 + x^2) = 25 + x^2 \implies 100 - 16\sqrt{25 + x^2} \implies \sqrt{25 + x^2} = \frac{25}{4} \implies 25 + x^2 = \frac{625}{16} \implies 400 + 16x^2 = 625 \implies 16x^2 = 225 \implies x = \frac{15}{4}</math>. Now, remember <math>CE \neq \frac{15}{4}</math>. <math>x = \frac{15}{4} = ME</math>, since we set <math>x = ME</math> in the start of the solution. Now to find the area <math>\frac{15}{4} \cdot 5 \cdot \frac{1}{2} = \frac{75}{8} = CE</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 19:09, 15 June 2017

Problem

Rectangle $ABCD$ has $AB=8$ and $BC=6$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$. What is the area of $\triangle AME$?

$\text{(A) } \frac{65}{8} \qquad \text{(B) } \frac{25}{3} \qquad \text{(C) } 9 \qquad \text{(D) } \frac{75}{8} \qquad \text{(E) } \frac{85}{8}$

Solution

Solution 1

[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",Ep,S); label("$M$",M,NW); [/asy]

By the Pythagorean theorem we have $AC=10$, hence $AM=5$.

The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar.

The ratio of their sides is $\frac{AM}{AB} = \frac 58$, hence the ratio of their areas is $\left( \frac 58 \right)^2 = \frac{25}{64}$.

And as the area of triangle $ABC$ is $\frac{6\cdot 8}2 = 24$, the area of triangle $AME$ is $24\cdot \frac{25}{64} = \boxed{ \frac{75}8 }$.

Solution 2 (Only Pythagorean Theorem)

[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); draw( C--Ep ); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",Ep,S); label("$M$",M,NW); [/asy]

Draw $EC$ as shown from the diagram. Since $AC$ is of length $10$, we have that $AM$ is of length $5$, because of the midpoint $M$. Through the Pythagorean theorem, we know that $AE^2 = AM^2 + ME^2 \implies 25 + ME^2$, which means $AE = \sqrt{25 + ME^2}$. Define $ME$ to be $x$ for the sake of clarity. We know that $EB = 8 - \sqrt{25 + x^2}$. From here, we know that $CE^2 = CB^2 + BE^2 = ME^2 + MC^2$. From here, we can write the expression $6^2 + (8 - \sqrt{25 + x^2})^2 = 5^2 + x^2 \implies 36 + (64 - 16\sqrt{25 + x^2} + 25 + x^2) = 25 + x^2 \implies 100 - 16\sqrt{25 + x^2} \implies \sqrt{25 + x^2} = \frac{25}{4} \implies 25 + x^2 = \frac{625}{16} \implies 400 + 16x^2 = 625 \implies 16x^2 = 225 \implies x = \frac{15}{4}$. Now, remember $CE \neq \frac{15}{4}$. $x = \frac{15}{4} = ME$, since we set $x = ME$ in the start of the solution. Now to find the area $\frac{15}{4} \cdot 5 \cdot \frac{1}{2} = \frac{75}{8} = CE$

See Also

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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