Difference between revisions of "2023 AMC 8 Problems/Problem 24"
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<asy> | <asy> | ||
//Diagram by TheMathGuyd | //Diagram by TheMathGuyd | ||
− | size( | + | size(12cm); |
real h = 2.5; // height | real h = 2.5; // height | ||
real g=4; //c2c space | real g=4; //c2c space | ||
Line 20: | Line 20: | ||
− | fill (shift(g,0)*(LTWO--B--cycle), | + | fill (shift(g,0)*(LTWO--B--cycle),mediumgrey); |
− | fill (LONE--A--C--cycle, | + | fill (LONE--A--C--cycle,mediumgrey); |
draw(LONE); | draw(LONE); | ||
Line 44: | Line 44: | ||
label("$11$", midpoint((0,h*(1-s))--B),UnFill); | label("$11$", midpoint((0,h*(1-s))--B),UnFill); | ||
</asy> | </asy> | ||
− | |||
<math>\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } 15.4</math> | <math>\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } 15.4</math> | ||
==Solution 1== | ==Solution 1== | ||
− | First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is <math>[ | + | First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is <math>[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)</math>. Similarly, we can find that the area of the gray part in the second triangle is <math>[ABC]\cdot\left(\tfrac{h-5}{h}\right)^2</math>. These areas are equal, so <math>1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2</math>. Simplifying yields <math>10h=146</math> so <math>h=\boxed{\textbf{(A) }14.6}</math>. |
~MathFun1000 (~edits apex304) | ~MathFun1000 (~edits apex304) | ||
− | ==Solution 2 ( | + | ==Solution 2 (Thorough)== |
− | We can call the length of AC as <math>x</math>. Therefore, the length of the base of the triangle with height <math>11</math> is <math>11/h = a/x</math>. Therefore, the base of the smaller triangle is <math>11x/h</math>. We find that the area of the trapezoid is <math>( | + | We can call the length of AC as <math>x</math>. Therefore, the length of the base of the triangle with height <math>11</math> is <math>11/h = a/x</math>. Therefore, the base of the smaller triangle is <math>11x/h</math>. We find that the area of the trapezoid is <math>(hx)/2 - 11^2x/2h</math>. |
− | Using similar triangles once again, we find that the base of the shaded triangle is <math>(h-5)/h = b/x</math>. Therefore, the area is <math>(h-5)(hx-5x)/h</math>. | + | Using similar triangles once again, we find that the base of the shaded triangle is <math>(h-5)/h = b/x</math>. Therefore, the area is <math>(h-5)(hx-5x)/h</math>. |
+ | |||
+ | Since the areas are the same, we find that <math>(hx)/2 - 121x/2h = (h-5)(hx-5x)/h</math>. Multiplying each side by <math>2h</math>, we get <math>h^2x - 121x = h^2x - 5hx - 5hx + 25x</math>. Therefore, we can subtract <math>25x + h^2x</math> from both sides, and get <math>-146x = -10hx</math>. Finally, we divide both sides by <math>-x</math> and get <math>10h = 146</math>. <math>h</math> is <math>\boxed{\textbf{(A)}14.6}</math>. | ||
+ | |||
+ | Solution by CHECKMATE2021 | ||
+ | |||
+ | ==Solution 3 (Faster)== | ||
+ | |||
+ | Since the length of AC does not matter, we can assume the base of triangle ABC is <math>h</math>. Therefore, the area of the trapezoid in the first diagram is <math>h^2/2 - \frac{11^2}{2}</math>. | ||
+ | |||
+ | The area of the triangle in the second diagram is now <math>\frac{(h-5)^2}{2}</math>. | ||
+ | |||
+ | Therefore, <math>\frac{h^2 - 11^2}{2} = \frac{(h-5)^2}{2}</math>. Multiplying both sides by <math>2</math>, we get <math>h^2 - 121 = h^2 - 10h + 25</math>. Subtracting <math>h^2 + 25</math> from both sides, we get <math>-146 = -10h</math> and <math>h</math> is <math>\boxed{\textbf{(A)}14.6}</math>. | ||
+ | |||
+ | Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]], and CHECKMATE2021 | ||
+ | |||
+ | ==Solution 4(Based on solution 3) (VERY SLOW)== | ||
+ | The answers are there on the bottom, so start with the middle one, <math>{\textbf{(C)}15}</math>. After calculating, we find that we need a shorter length, so try <math>{\textbf{(B)}14.8}</math>. Still, we need a shorter answer, so we simply choose <math>\boxed{\textbf{(A)}14.6}</math> without trying it out. | ||
+ | |||
+ | ~SaxStreak | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/Ku_c1YHnLt0?si=NHwA1x9STOJLG8IP&t=5349 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | https://youtu.be/uxyJGZ3ZYGE | ||
+ | |||
+ | ~please like and subscribe | ||
+ | |||
+ | ==Video Solution (THINKING CREATIVELY!!!)== | ||
+ | |||
+ | https://youtu.be/SVVSMcw1Xe8 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution 1 by OmegaLearn (Using Similarity)== | ==Video Solution 1 by OmegaLearn (Using Similarity)== | ||
Line 65: | Line 100: | ||
https://www.youtube.com/watch?v=GTlkTwxSxgo | https://www.youtube.com/watch?v=GTlkTwxSxgo | ||
− | ==Video Solution 3 by Magic Square (Using Similarity and Special Value)== | + | ==Video Solution 3 by Magic Square (Using Similarity and Special Value)(best solution)== |
https://www.youtube.com/watch?v=-N46BeEKaCQ&t=1569s | https://www.youtube.com/watch?v=-N46BeEKaCQ&t=1569s | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/DBqko2xATxs&t=3270 | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/roTSeCAehek | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by harungurcan== | ||
+ | https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1593s | ||
+ | |||
+ | ~harungurcan | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=23|num-a=25}} | {{AMC8 box|year=2023|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:10, 17 June 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Thorough)
- 4 Solution 3 (Faster)
- 5 Solution 4(Based on solution 3) (VERY SLOW)
- 6 Video Solution
- 7 Video Solution by Math-X (First understand the problem!!!)
- 8 Video Solution (THINKING CREATIVELY!!!)
- 9 Video Solution 1 by OmegaLearn (Using Similarity)
- 10 Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)
- 11 Video Solution 3 by Magic Square (Using Similarity and Special Value)(best solution)
- 12 Video Solution by Interstigation
- 13 Video Solution by WhyMath
- 14 Video Solution by harungurcan
- 15 See Also
Problem
Isosceles has equal side lengths and . In the figure below, segments are drawn parallel to so that the shaded portions of have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of of ?
Solution 1
First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is . Similarly, we can find that the area of the gray part in the second triangle is . These areas are equal, so . Simplifying yields so .
~MathFun1000 (~edits apex304)
Solution 2 (Thorough)
We can call the length of AC as . Therefore, the length of the base of the triangle with height is . Therefore, the base of the smaller triangle is . We find that the area of the trapezoid is .
Using similar triangles once again, we find that the base of the shaded triangle is . Therefore, the area is .
Since the areas are the same, we find that . Multiplying each side by , we get . Therefore, we can subtract from both sides, and get . Finally, we divide both sides by and get . is .
Solution by CHECKMATE2021
Solution 3 (Faster)
Since the length of AC does not matter, we can assume the base of triangle ABC is . Therefore, the area of the trapezoid in the first diagram is .
The area of the triangle in the second diagram is now .
Therefore, . Multiplying both sides by , we get . Subtracting from both sides, we get and is .
Solution by ILoveMath31415926535, and CHECKMATE2021
Solution 4(Based on solution 3) (VERY SLOW)
The answers are there on the bottom, so start with the middle one, . After calculating, we find that we need a shorter length, so try . Still, we need a shorter answer, so we simply choose without trying it out.
~SaxStreak
Video Solution
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/Ku_c1YHnLt0?si=NHwA1x9STOJLG8IP&t=5349
~Math-X
~please like and subscribe
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution 1 by OmegaLearn (Using Similarity)
Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)
https://www.youtube.com/watch?v=GTlkTwxSxgo
Video Solution 3 by Magic Square (Using Similarity and Special Value)(best solution)
https://www.youtube.com/watch?v=-N46BeEKaCQ&t=1569s
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=3270
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1593s
~harungurcan
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.