Difference between revisions of "2023 AMC 8 Problems/Problem 24"
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==Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)== | ==Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)== | ||
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)== | ||
| + | https://www.youtube.com/watch?v=-N46BeEKaCQ&t=1569s | ||
==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 10:16, 25 January 2023
Contents
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 ?
![[asy] //Diagram by TheMathGuyd size(9cm); real h = 2.5; // height real g=4; //c2c space real s = 0.65; //Xcord of Hline real adj = 0.08; //adjust line diffs pair A,B,C; B=(0,h); C=(1,0); A=-conj(C); pair PONE=(s,h*(1-s)); //Endpoint of Hline ONE pair PTWO=(s+adj,h*(1-s-adj)); //Endpoint of Hline ONE path LONE=PONE--(-conj(PONE)); //Hline ONE path LTWO=PTWO--(-conj(PTWO)); path T=A--B--C--cycle; //Triangle fill (shift(g,0)*(LTWO--B--cycle),grey); fill (LONE--A--C--cycle,grey); draw(LONE); draw(T); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); draw(shift(g,0)*LTWO); draw(shift(g,0)*T); label("$A$",shift(g,0)*A,SW); label("$B$",shift(g,0)*B,N); label("$C$",shift(g,0)*C,SE); draw(B--shift(g,0)*B,dashed); draw(C--shift(g,0)*A,dashed); draw((g/2,0)--(g/2,h),dashed); draw((0,h*(1-s))--B,dashed); draw((g,h*(1-s-adj))--(g,0),dashed); label("$5$", midpoint((g,h*(1-s-adj))--(g,0)),UnFill); label("$h$", midpoint((g/2,0)--(g/2,h)),UnFill); label("$11$", midpoint((0,h*(1-s))--B),UnFill); [/asy]](http://latex.artofproblemsolving.com/1/f/6/1f630ff2f2b0c0154af15960436fc1c68b5a7a0e.png)
(note: diagrams are not necessarily 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 ![$[\text{ABC}]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$](http://latex.artofproblemsolving.com/c/c/d/ccd19fe2d21c98e12c3f7a6e1f715aee46085b3d.png)
. Similarly, we can find that the area of the gray part in the second triangle is ![$[\text{ABC}]\cdot\left(\tfrac{h-5}{h}\right)^2$](http://latex.artofproblemsolving.com/0/1/c/01c770040e946593b2026999898568707d5b5538.png)
. These areas are equal, so 
. Simplifying yields 
so 
.
~MathFun1000 (~edits apex304)
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)
https://www.youtube.com/watch?v=-N46BeEKaCQ&t=1569s
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. 
