Difference between revisions of "2023 AMC 8 Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | Isosceles <math>\triangle ABC</math> has equal side lengths <math>AB</math> and <math>BC</math>. In the figure below, segments are drawn parallel to <math>\overline{AC}</math> so that the shaded portions of <math>\triangle ABC</math> have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of <math>h</math> of <math>\triangle ABC</math>? | + | Isosceles <math>\triangle ABC</math> has equal side lengths <math>AB</math> and <math>BC</math>. In the figure below, segments are drawn parallel to <math>\overline{AC}</math> so that the shaded portions of <math>\triangle ABC</math> have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of <math>h</math> of <math>\triangle ABC</math>? (Diagram not drawn to scale.) |
<asy> | <asy> | ||
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> | ||
Line 53: | Line 52: | ||
~MathFun1000 (~edits apex304) | ~MathFun1000 (~edits apex304) | ||
− | ==Solution 2 | + | ==Solution 2== |
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>. | 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)/ | + | 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)/2h</math>. |
− | Since the areas are the same, we find that <math>(hx)/2 - 121x/2h = (h-5)(hx-5x)/ | + | Since the areas are the same, we find that <math>(hx)/2 - 121x/2h = (h-5)(hx-5x)/2h</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 | + | Solution by CHECKMATE2021 (edits by Someone) |
==Solution 3 (Faster)== | ==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 - 11^2 | + | 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>(h-5)^2 | + | The area of the triangle in the second diagram is now <math>\frac{(h-5)^2}{2}</math>. |
− | Therefore, <math>h^ | + | 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]] | + | Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]], and CHECKMATE2021 |
− | ==Solution | + | ==Video Solution (A Clever Explanation You’ll Get Instantly)== |
− | + | https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=2990 | |
+ | ~hsnacademy | ||
− | + | ==Video Solution by Math-X== | |
+ | https://youtu.be/Ku_c1YHnLt0?si=NHwA1x9STOJLG8IP&t=5349 | ||
− | ==Video Solution 1 by OmegaLearn (Using | + | ~Math-X |
+ | |||
+ | https://youtu.be/uxyJGZ3ZYGE | ||
+ | |||
+ | ==Video Solution (THINKING CREATIVELY!!!)== | ||
+ | |||
+ | https://youtu.be/SVVSMcw1Xe8 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution 1 by OmegaLearn (Using Similar Triangles)== | ||
https://youtu.be/almtw4n-92A | https://youtu.be/almtw4n-92A | ||
Line 84: | Line 95: | ||
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== | ==Video Solution by Interstigation== | ||
− | https://youtu.be/ | + | https://youtu.be/DBqko2xATxs&t=3270 |
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Line 98: | Line 110: | ||
~harungurcan | ~harungurcan | ||
+ | |||
+ | ==Video Solution by Dr. David== | ||
+ | https://youtu.be/GnlU-McyPXY | ||
==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}} |
Latest revision as of 14:29, 19 January 2025
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Faster)
- 5 Video Solution (A Clever Explanation You’ll Get Instantly)
- 6 Video Solution by Math-X
- 7 Video Solution (THINKING CREATIVELY!!!)
- 8 Video Solution 1 by OmegaLearn (Using Similar Triangles)
- 9 Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)
- 10 Video Solution 3 by Magic Square (Using Similarity and Special Value)(best solution)
- 11 Video Solution by Interstigation
- 12 Video Solution by WhyMath
- 13 Video Solution by harungurcan
- 14 Video Solution by Dr. David
- 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
? (Diagram not drawn to scale.)
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
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 (edits by Someone)
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
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=2990 ~hsnacademy
Video Solution by Math-X
https://youtu.be/Ku_c1YHnLt0?si=NHwA1x9STOJLG8IP&t=5349
~Math-X
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution 1 by OmegaLearn (Using Similar Triangles)
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
Video Solution by Dr. David
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.