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Difference between revisions of "2024 AMC 10B Problems/Problem 11"

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~pepper2831  
 
~pepper2831  
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==Solution 4 (Similar Triangles)==
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We are given <math>WX = 4</math>, <math>WZ = 8</math>. △ WXM and △ MYA have equal area, so let <math>XM = 2x</math> and <math>ZA = x</math>. <math>MY = 8-2x</math> and <math>AY = 4-x</math>.
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From this, we can conclude that <math>\frac{MY}{AY} = \frac{8-2x}{4-x} = \frac{2}{1}</math>
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Since <math>WM</math> intersects parallel lines <math>WZ</math> and <math>XY</math>, <math>\angle ZWM = \angle WMZ</math>. <math>\angle ZWM + \angle MWX = 90^\circ</math>, so <math>180^\circ - 90^\circ = \angle WMZ + \angle AMY</math>. Thus, <math>\angle MWX = \angle AMY</math> and △ WXM ~ △ MYA due to AA Similarity.
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Corresponding sides of similar triangles are proportional, so <math>\frac{WX}{XM} = \frac{MY}{AY}</math> or <math>\frac{4}{2x} = \frac{2}{1}</math>. It is clear that <math>2x = 2</math>, and <math>x = 1</math>. Now, all we have to do is subtract the area of the rectangle by each of the three triangles.
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△ WMA = <math>8</math> · <math>4</math> - (<math>\frac{1}{2}</math> · <math>4</math> · <math>2</math>) - (<math>\frac{1}{2}</math> · <math>8</math> · <math>1</math>) - (<math>\frac{1}{2}</math> · <math>6</math> · <math>3</math>)
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△ WMA = <math>32 - 4 - 4 - 9</math>
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△ WMA = <math>\boxed{\textbf{(C) }15}</math>
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~peeghj
  
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

Revision as of 21:39, 14 November 2024

The following problem is from both the 2024 AMC 10B #11 and 2024 AMC 12B #7, so both problems redirect to this page.

Problem

In the figure below $WXYZ$ is a rectangle with $WX=4$ and $WZ=8$. Point $M$ lies $\overline{XY}$, point $A$ lies on $\overline{YZ}$, and $\angle WMA$ is a right angle. The areas of $\triangle WXM$ and $\triangle WAZ$ are equal. What is the area of $\triangle WMA$?

[asy] pair X = (0, 0); pair W = (0, 4); pair Y = (8, 0); pair Z = (8, 4); label("$X$", X, dir(180)); label("$W$", W, dir(180)); label("$Y$", Y, dir(0)); label("$Z$", Z, dir(0)); draw(W--X--Y--Z--cycle); dot(X); dot(Y); dot(W); dot(Z); pair M = (2, 0); pair A = (8, 3); label("$A$", A, dir(0)); dot(M); dot(A); draw(W--M--A--cycle); markscalefactor = 0.05; draw(rightanglemark(W, M, A)); label("$M$", M, dir(-90)); [/asy]

$\textbf{(A) }13 \qquad \textbf{(B) }14 \qquad \textbf{(C) }15 \qquad \textbf{(D) }16 \qquad \textbf{(E) }17 \qquad$

Solution

Solution 1

We know that $WX = 4$, $WZ = 8$, so $YZ = 4$ and $YX = 8$. Since $\angle WMA = 90^\circ$, triangles $WXM$ and $MYA$ are similar. Therefore, $\frac{WX}{MY} = \frac{XM}{YA}$, which gives $\frac{4}{8 - XM} = \frac{XM}{4 - ZA}$. We also know that the areas of triangles $WXM$ and $WAZ$ are equal, so $WX \cdot XM = WZ \cdot ZA$, which implies $4 \cdot XM = 8 \cdot ZA$. Substituting this into the previous equation, we get $\frac{4}{8 - 2ZA} = \frac{XM}{4 - ZA}$, yielding $ZA = 1$ and $XM = 2$. Thus,

\[\triangle WMA = 4 \cdot 8 - \frac{4 \cdot 2}{2} - \frac{8 \cdot 1}{2} - \frac{6 \cdot 3}{2} = \boxed{\textbf{(C) }15}\]

~Athmyx

Solution 2

Let $XM=b$, $ZA = a$, $4\cdot b= 8\cdot a$, $b = 2a$, \[WM^2 + AM^2 = AW^2\] \[(b^2+4^2) + (4-a)^2 + (8-b)^2 = (a^2 + 8^2)\] a=1, b=2 , \[\triangle WMA = area(WXYZ) - \triangle WZA- \triangle WXM- \triangle MYA = 32 - 4-4-9 = \boxed{\textbf{(C) }15}\]

~luckuso ~minor edits by EaZ_Shadow

Solution 3 (Pythagorean Theorem)

Assign ZA as $x$, then AY as $4 - x$. Assign XM as $y$ and MY as $8 - y$. Since triangles WXM and WZA are together, we can say $4x = 8y$, so $y = 2x$. Then therefore, XM is $2x$ and MY has length $8 - 2x$. We can use the Pythagorean theorem to find WM, which is actually $\sqrt{(2x)^2 + 4^2)} = \sqrt{4x^2 + 16}$. We don't factor it yet - we are going to find $x$ again using the Pythagorean Theorem. Similarly, finding MA is just the square root of the squares of AY and MY individually, or $\sqrt{(8 - 2x)^2 + (4 - x)^2} = \sqrt{64 - 32x + 4x^2 + 16 - 8x + x^2} = \sqrt{5x^2 - 40x + 80}$. Then simply, WA is really $\sqrt{x^2 + 64}$.

Now we have the three sides of the right triangle: $\sqrt{4x^2 + 16}$, $\sqrt{5x^2 - 40x + 80}$, and $\sqrt{x^2 + 64}$. Per the Pythagorean theorem again, we can see $(4x^2 + 16) + (5x^2 - 40x + 80) = (x^2 + 64)$. Combining like terms gives us $8x^2 - 40x + 32 = 0$, then dividing by 8 gives $x^2 - 5x + 4 = 0$. As this elementary and well-known quadratic gives us the roots of $1$ and $4$, we can see it is a bit weird to have $x = 4$, as then point Z is point A. So we'll assume $x = 1$. We have two legs of the triangle by plugging in the sides with x in them, given that $x = 1$: $\sqrt{20}$ and $\sqrt{45}$. We should know that $20 \cdot 45 = 900$, and $\sqrt{900} = 30.$ Dividing by 2 reveals us our answer: $\boxed{\textbf{(C) }15}$

~pepper2831

Solution 4 (Similar Triangles)

We are given $WX = 4$, $WZ = 8$. △ WXM and △ MYA have equal area, so let $XM = 2x$ and $ZA = x$. $MY = 8-2x$ and $AY = 4-x$. From this, we can conclude that $\frac{MY}{AY} = \frac{8-2x}{4-x} = \frac{2}{1}$

Since $WM$ intersects parallel lines $WZ$ and $XY$, $\angle ZWM = \angle WMZ$. $\angle ZWM + \angle MWX = 90^\circ$, so $180^\circ - 90^\circ = \angle WMZ + \angle AMY$. Thus, $\angle MWX = \angle AMY$ and △ WXM ~ △ MYA due to AA Similarity.

Corresponding sides of similar triangles are proportional, so $\frac{WX}{XM} = \frac{MY}{AY}$ or $\frac{4}{2x} = \frac{2}{1}$. It is clear that $2x = 2$, and $x = 1$. Now, all we have to do is subtract the area of the rectangle by each of the three triangles.


△ WMA = $8$ · $4$ - ($\frac{1}{2}$ · $4$ · $2$) - ($\frac{1}{2}$ · $8$ · $1$) - ($\frac{1}{2}$ · $6$ · $3$)

△ WMA = $32 - 4 - 4 - 9$

△ WMA = $\boxed{\textbf{(C) }15}$

~peeghj

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/YqKmvSR1Ckk?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE


See also

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 12 Problems and Solutions

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

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