Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AMC 10B Problems/Problem 18"

(→‎Solution 2)
m (→‎Solution 5 (Pythagorean Theorem))
 
(14 intermediate revisions by 7 users not shown)
Line 14: Line 14:
 
</math>
 
</math>
  
−
== Solution 1 (Coordinate Geo)==
+
== Solution 1 (Coordinate Geometry)==
  
−
Set <math>A</math> to <math>(0,0)</math>. Since <math>M</math> is the midpoint of the diagonal, it would be <math>(4,-3)</math>. The diagonal <math>AC</math> would be the line <math>y = -\frac{3x}{4}</math>. Since <math>ME</math> is perpendicular to <math>AC</math>, its line would be in the form <math>y = \frac{4x}{3} + b</math>. Plugging in <math>4</math> and <math>-3</math> for <math>x</math> and <math>y</math> would give <math>b = \frac{25}{3}</math>. To find the x-intercept of <math>y = \frac{4x}{3} + \frac{25}{3}</math> we plug in <math>0</math> for <math>y</math> and get <math>x = \frac{25}{4}</math>. Then, using the Shoelace Formula for <math>(0,0)</math> , <math>(4,-3)</math>, and <math>(\frac{25}{4}, 0)</math>, we find the area is <math>\frac{75}{8}</math>.
+
Set <math>A</math> to <math>(0,0)</math>. Since <math>M</math> is the midpoint of the diagonal, it would be <math>(4,3)</math>. The diagonal <math>AC</math> would be the line <math>y = \frac{3x}{4}</math>. Since <math>ME</math> is perpendicular to <math>AC</math>, its line would be in the form <math>y = -\frac{4x}{3} + b</math>. Plugging in <math>4</math> and <math>3</math> for <math>x</math> and <math>y</math> would give <math>b = \frac{25}{3}</math>. To find the x-intercept of <math>y = -\frac{4x}{3} + \frac{25}{3}</math> we plug in <math>0</math> for <math>y</math> and get <math>x = \frac{25}{4}</math>. Then, using the Shoelace Formula for <math>(0,0)</math> , <math>(4,3)</math>, and <math>(\frac{25}{4}, 0)</math>, we find the area is <math>\frac{75}{8}</math>.
  
 
== Solution 2 ==  
 
== Solution 2 ==  
Line 25: Line 25:
 
path ortho = shift(M)*rotate(-90)*(A--C);
 
path ortho = shift(M)*rotate(-90)*(A--C);
 
pair Ep = intersectionpoint(ortho, A--B);
 
pair Ep = intersectionpoint(ortho, A--B);
−
draw( A--B--C--Dsxklskl--cycle );
+
draw( A--B--C--D--cycle );
 
draw( A--C );
 
draw( A--C );
 
draw( M--Ep );
 
draw( M--Ep );
Line 70: Line 70:
 
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>
 
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>
  
 +
==Solution 4 (Similarity) ==
 +
We know <math>AM = \frac{10}{2} = 5</math> by the Pythagorean theorem, and furthermore, <math>\triangle AME</math> is similar to <math>\triangle ABC</math>. Therefore, <math>ME = 5 \cdot \frac{6}{8} = \frac{15}{4}</math>, and the area of the triangle is <math>5 \cdot \frac{15}{4} \cdot \frac{1}{2} = \boxed{\frac{75}{8}}</math>.
 +
 +
==Solution 5 (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>
 +
 +
By the [[Pythagorean Theorem]], we claim that <math>AC = 10</math>. It then follows that <math>AM \cong MC = 5.</math>
  
−
==Video Solution==
+
Because we have <math>AM \cong MC, \angle AME \cong \angle CME,</math> and reflexive side <math>EM</math>, it follows that <math>\triangle AME \cong \triangle CME.</math> By CPCTC, we have <math>AE \cong EC.</math> For the sake of simplicity, we'll call those side lengths <math>x</math>. Also, since <math>AE = x,</math> we get <math>BE = 8 - x.</math> We can now set up the Pythagorean theorem on <math>\triangle EBC</math>: <cmath>(8 - x)^2 + 6^2 = x^2.</cmath> Combining like terms and simplifying gives <math>-16x + 100 = 0</math> so <math>x = \frac{25}{4}.</math>
−
https://www.youtube.com/watch?v=dQw4w9WgXcQ
 
  
−
~Anonymous23
+
It helps to think that in order to find <math>[AME],</math> we must have <math>\overline{MC}</math> and <math>\overline{EM}.</math> Let <math>EM = y.</math> Applying the Pythagorean Theorem to <math>\triangle CME</math> gives <cmath>5^2 + y^2 = \left(\frac{25}{4} \right)^2.</cmath> Solving for <math>y</math> (this is not that difficult) gives <math>y = \frac{15}{4}.</math> So, the area of <math>\triangle AME</math> is <math>\frac{\frac{15}{4} \cdot 5}{2} = \frac{75}{8} \implies \boxed{D}.</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 21:14, 18 January 2024

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 1 (Coordinate Geometry)

Set $A$ to $(0,0)$. Since $M$ is the midpoint of the diagonal, it would be $(4,3)$. The diagonal $AC$ would be the line $y = \frac{3x}{4}$. Since $ME$ is perpendicular to $AC$, its line would be in the form $y = -\frac{4x}{3} + b$. Plugging in $4$ and $3$ for $x$ and $y$ would give $b = \frac{25}{3}$. To find the x-intercept of $y = -\frac{4x}{3} + \frac{25}{3}$ we plug in $0$ for $y$ and get $x = \frac{25}{4}$. Then, using the Shoelace Formula for $(0,0)$ , $(4,3)$, and $(\frac{25}{4}, 0)$, we find the area is $\frac{75}{8}$.

Solution 2

[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]

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 3 (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 $\frac{15}{4} \cdot 5 \cdot \frac{1}{2} = \frac{75}{8} = CE$

Solution 4 (Similarity)

We know $AM = \frac{10}{2} = 5$ by the Pythagorean theorem, and furthermore, $\triangle AME$ is similar to $\triangle ABC$. Therefore, $ME = 5 \cdot \frac{6}{8} = \frac{15}{4}$, and the area of the triangle is $5 \cdot \frac{15}{4} \cdot \frac{1}{2} = \boxed{\frac{75}{8}}$.

Solution 5 (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]

By the Pythagorean Theorem, we claim that $AC = 10$. It then follows that $AM \cong MC = 5.$

Because we have $AM \cong MC, \angle AME \cong \angle CME,$ and reflexive side $EM$, it follows that $\triangle AME \cong \triangle CME.$ By CPCTC, we have $AE \cong EC.$ For the sake of simplicity, we'll call those side lengths $x$. Also, since $AE = x,$ we get $BE = 8 - x.$ We can now set up the Pythagorean theorem on $\triangle EBC$: \[(8 - x)^2 + 6^2 = x^2.\] Combining like terms and simplifying gives $-16x + 100 = 0$ so $x = \frac{25}{4}.$

It helps to think that in order to find $[AME],$ we must have $\overline{MC}$ and $\overline{EM}.$ Let $EM = y.$ Applying the Pythagorean Theorem to $\triangle CME$ gives \[5^2 + y^2 = \left(\frac{25}{4} \right)^2.\] Solving for $y$ (this is not that difficult) gives $y = \frac{15}{4}.$ So, the area of $\triangle AME$ is $\frac{\frac{15}{4} \cdot 5}{2} = \frac{75}{8} \implies \boxed{D}.$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_18&oldid=211222"