Difference between revisions of "2023 AMC 8 Problems/Problem 24"
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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>? | ||
− | * | + | <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> | ||
(note: diagrams are not necessarily drawn to scale) | (note: diagrams are not necessarily drawn to scale) | ||
Revision as of 03:31, 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
?
(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 . 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)
Video Solution 1 by OmegaLearn (Using Similarity)
Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)
https://www.youtube.com/watch?v=GTlkTwxSxgo
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.