Difference between revisions of "2007 AMC 10B Problems/Problem 21"
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[[Image:2007AMC10B21.png|center]] | [[Image:2007AMC10B21.png|center]] | ||
− | There are | + | There are lots of similar triangles in the diagram, but we will only use <math>\triangle WBZ \sim \triangle ABC.</math> If <math>h</math> is the altitude from <math>B</math> to <math>AC</math> and <math>s</math> is the sidelength of the square, then <math>h-s</math> is the altitude from <math>B</math> to <math>WZ.</math> By similar triangles, |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \frac{h-s}{ | + | \frac{h-s}{h}&=\frac{s}{5}\\ |
+ | 5(h-s)&=hs\\ | ||
5h-5s&=hs\\ | 5h-5s&=hs\\ | ||
5h&=s(h+5)\\ | 5h&=s(h+5)\\ | ||
Line 34: | Line 35: | ||
<cmath>\frac{ZB}{l} = \frac{4}{5} \rightarrow ZB = \frac{4}{5}l</cmath> | <cmath>\frac{ZB}{l} = \frac{4}{5} \rightarrow ZB = \frac{4}{5}l</cmath> | ||
− | But then <math>\frac{5}{3}l+\frac{4}{5}l = CZ+ZB = CB = 4 \longrightarrow \frac{37}{15}l=4 \longrightarrow l = \frac{60}{37} \Longrightarrow \mathrm{( | + | But then <math>\frac{5}{3}l+\frac{4}{5}l = CZ+ZB = CB = 4 \longrightarrow \frac{37}{15}l=4 \longrightarrow l = \frac{60}{37} \Longrightarrow \boxed{\mathrm{(B)}\frac{60}{37}}</math> |
+ | == Video Solution by OmegaLearn== | ||
+ | https://youtu.be/FDgcLW4frg8?t=4662 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 16:52, 29 July 2024
Problem
Right has
and
Square
is inscribed in
with
and
on
on
and
on
What is the side length of the square?
Solution 1
There are lots of similar triangles in the diagram, but we will only use If
is the altitude from
to
and
is the sidelength of the square, then
is the altitude from
to
By similar triangles,
Find the length of the altitude of Since it is a right triangle, the area of
is
The area can also be expressed as so
Substitute back into
Solution 2
Let be the side length of the inscribed square. Note that
.
Then we can setup the following ratios:
But then
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=4662
~ pi_is_3.14
See Also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.