2023 AMC 8 Problems/Problem 24: Difference between revisions
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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 02:31, 25 January 2023
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing