Difference between revisions of "2024 AMC 10B Problems/Problem 11"
Pepper2831 (talk | contribs) |
|||
(15 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
− | "1 | + | {{duplicate|[[2024 AMC 10B Problems/Problem 11|2024 AMC 10B #11]] and [[2024 AMC 12B Problems/Problem 7|2024 AMC 12B #7]]}} |
− | https://www. | + | |
− | + | ==Problem== | |
+ | In the figure below <math>WXYZ</math> is a rectangle with <math>WX=4</math> and <math>WZ=8</math>. Point <math>M</math> lies <math>\overline{XY}</math>, point <math>A</math> lies on <math>\overline{YZ}</math>, and <math>\angle WMA</math> is a right angle. The areas of <math>\triangle WXM</math> and <math>\triangle WAZ</math> are equal. What is the area of <math>\triangle WMA</math>? | ||
+ | |||
+ | <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> | ||
+ | |||
+ | Note: On certain tests that took place in China, the problem asked for the area of <math>\triangle MAY</math>. | ||
+ | |||
+ | <math> | ||
+ | \textbf{(A) }13 \qquad | ||
+ | \textbf{(B) }14 \qquad | ||
+ | \textbf{(C) }15 \qquad | ||
+ | \textbf{(D) }16 \qquad | ||
+ | \textbf{(E) }17 \qquad | ||
+ | </math> | ||
+ | |||
+ | [[2024 AMC 12B Problems/Problem 7|Solution]] | ||
+ | |||
+ | ==Solution 1== | ||
+ | We know that <math>WX = 4</math>, <math>WZ = 8</math>, so <math>YZ = 4</math> and <math>YX = 8</math>. Since <math>\angle WMA = 90^\circ</math>, triangles <math>WXM</math> and <math>MYA</math> are similar. Therefore, <math>\frac{WX}{MY} = \frac{XM}{YA}</math>, which gives <math>\frac{4}{8 - XM} = \frac{XM}{4 - ZA}</math>. We also know that the areas of triangles <math>WXM</math> and <math>WAZ</math> are equal, so <math>WX \cdot XM = WZ \cdot ZA</math>, which implies <math>4 \cdot XM = 8 \cdot ZA</math>. Substituting this into the previous equation, we get <math>\frac{4}{8 - 2ZA} = \frac{XM}{4 - ZA}</math>, yielding <math>ZA = 1</math> and <math>XM = 2</math>. Thus, | ||
+ | |||
+ | <cmath> | ||
+ | \triangle WMA = 4 \cdot 8 - \frac{4 \cdot 2}{2} - \frac{8 \cdot 1}{2} - \frac{6 \cdot 3}{2} = \boxed{\textbf{(C) }15} | ||
+ | </cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx] | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>XM=b</math>, <math>ZA = a</math>, <math>4\cdot b= 8\cdot a</math>, <math>b = 2a</math>, | ||
+ | <cmath>WM^2 + AM^2 = AW^2</cmath> | ||
+ | <cmath>(b^2+4^2) + (4-a)^2 + (8-b)^2 = (a^2 + 8^2)</cmath> | ||
+ | a=1, b=2 , | ||
+ | <cmath> | ||
+ | \triangle WMA = area(WXYZ) - \triangle WZA- \triangle WXM- \triangle MYA = 32 - 4-4-9 = \boxed{\textbf{(C) }15} | ||
+ | </cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | ~minor edits by EaZ_Shadow | ||
+ | |||
+ | ==Solution 3 (Pythagorean Theorem) == | ||
+ | Assign ZA as <math>x</math>, then AY as <math>4 - x</math>. Assign XM as <math>y</math> and MY as <math>8 - y</math>. Since triangles WXM and WZA are together, we can say <math>4x = 8y</math>, so <math>y = 2x</math>. Then therefore, XM is <math>2x</math> and MY has length <math>8 - 2x</math>. We can use the Pythagorean theorem to find WM, which is actually <math>\sqrt{(2x)^2 + 4^2)} = \sqrt{4x^2 + 16}</math>. We don't factor it yet - we are going to find <math>x</math> again using the Pythagorean Theorem. Similarly, finding MA is just the square root of the squares of AY and MY individually, or <math>\sqrt{(8 - 2x)^2 + (4 - x)^2} = \sqrt{64 - 32x + 4x^2 + 16 - 8x + x^2} = \sqrt{5x^2 - 40x + 80}</math>. Then simply, WA is really <math>\sqrt{x^2 + 64}</math>. | ||
+ | |||
+ | Now we have the three sides of the right triangle: <math>\sqrt{4x^2 + 16}</math>, <math>\sqrt{5x^2 - 40x + 80}</math>, and <math>\sqrt{x^2 + 64}</math>. Per the Pythagorean theorem again, we can see <math>(4x^2 + 16) + (5x^2 - 40x + 80) = (x^2 + 64)</math>. Combining like terms gives us <math>8x^2 - 40x + 32 = 0</math>, then dividing by 8 gives <math>x^2 - 5x + 4 = 0</math>. As this elementary and well-known quadratic gives us the roots of <math>1</math> and <math>4</math>, we can see it is a bit weird to have <math>x = 4</math>, as then point Z is point A. So we'll assume <math>x = 1</math>. We have two legs of the triangle by plugging in the sides with x in them, given that <math>x = 1</math>: <math>\sqrt{20}</math> and <math>\sqrt{45}</math>. We should know that <math>20 \cdot 45 = 900</math>, and <math>\sqrt{900} = 30.</math> Dividing by 2 reveals us our answer: <math>\boxed{\textbf{(C) }15}</math> | ||
+ | |||
+ | ~pepper2831 | ||
+ | |||
+ | ==Solution 4 (Similar Triangles)== | ||
+ | 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>. | ||
+ | From this, we can conclude that <math>\frac{MY}{AY} = \frac{8-2x}{4-x} = \frac{2}{1}</math> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | |||
+ | △ 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>) | ||
+ | |||
+ | △ WMA = <math>32 - 4 - 4 - 9</math> | ||
+ | |||
+ | △ WMA = <math>\boxed{\textbf{(C) }15}</math> | ||
+ | |||
+ | ~peeghj | ||
+ | |||
+ | ==China Test Solution (Finding MAY)== | ||
+ | From solution 3, instead of finding <math>WMA</math>, we instead find MAY. Then <math>x = 1</math> then we have <math>MA = 8 - 2x = 6</math>. Again, since <math>AY = 4 - x</math>, then <math>AY = 4 - 1 = 3.</math> The area of a triangle with legs <math>3</math> and <math>6</math> is <math>\frac{3 * 6}{2} = \boxed{9}</math>. | ||
+ | |||
+ | ~pepper2831 (again) | ||
+ | |||
+ | |||
+ | ==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== | ||
+ | {{AMC10 box|year=2024|ab=B|num-b=10|num-a=12}} | ||
+ | {{AMC12 box|year=2024|ab=B|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Revision as of 18:17, 15 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.
Contents
[hide]Problem
In the figure below is a rectangle with and . Point lies , point lies on , and is a right angle. The areas of and are equal. What is the area of ?
Note: On certain tests that took place in China, the problem asked for the area of .
Solution 1
We know that , , so and . Since , triangles and are similar. Therefore, , which gives . We also know that the areas of triangles and are equal, so , which implies . Substituting this into the previous equation, we get , yielding and . Thus,
Solution 2
Let , , , , a=1, b=2 ,
~luckuso ~minor edits by EaZ_Shadow
Solution 3 (Pythagorean Theorem)
Assign ZA as , then AY as . Assign XM as and MY as . Since triangles WXM and WZA are together, we can say , so . Then therefore, XM is and MY has length . We can use the Pythagorean theorem to find WM, which is actually . We don't factor it yet - we are going to find again using the Pythagorean Theorem. Similarly, finding MA is just the square root of the squares of AY and MY individually, or . Then simply, WA is really .
Now we have the three sides of the right triangle: , , and . Per the Pythagorean theorem again, we can see . Combining like terms gives us , then dividing by 8 gives . As this elementary and well-known quadratic gives us the roots of and , we can see it is a bit weird to have , as then point Z is point A. So we'll assume . We have two legs of the triangle by plugging in the sides with x in them, given that : and . We should know that , and Dividing by 2 reveals us our answer:
~pepper2831
Solution 4 (Similar Triangles)
We are given , . △ WXM and △ MYA have equal area, so let and . and . From this, we can conclude that
Since intersects parallel lines and , . , so . Thus, and △ WXM ~ △ MYA due to AA Similarity.
Corresponding sides of similar triangles are proportional, so or . It is clear that , and . Now, all we have to do is subtract the area of the rectangle by each of the three triangles.
△ WMA = · - ( · · ) - ( · · ) - ( · · )
△ WMA =
△ WMA =
~peeghj
China Test Solution (Finding MAY)
From solution 3, instead of finding , we instead find MAY. Then then we have . Again, since , then The area of a triangle with legs and is .
~pepper2831 (again)
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.