Difference between revisions of "2023 AMC 10A Problems/Problem 11"
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<math>\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1</math> | <math>\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1</math> | ||
| − | ==Solution 1== | + | ==Solution== |
| + | |||
| + | Note that each side length is <math>\sqrt{2}</math> and <math>\sqrt{3}.</math> Let the shorter side of our triangle be <math>x</math>, thus the longer leg is <math>\sqrt{3}-x</math>. Hence, by the Pythagorean Theorem, we have <cmath>(x-\sqrt{3})^2+x^2=2</cmath> | ||
| + | <cmath>2x^2-2x\sqrt{3}+1=0</cmath>. | ||
| + | |||
| + | By the quadratic formula, we find <math>x=\frac{\sqrt{3}\pm1}{2}</math>. Hence, our answer is <math>\frac{\sqrt{3}-1}{\sqrt{3}+1}}=2-\sqrt{3}.</math> | ||
| + | |||
| + | ~SirAppel | ||
Revision as of 20:51, 9 November 2023
A square of area 
is inscribed in a square of area 
, creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?
[asy]
size(200);
defaultpen(linewidth(0.6pt)+fontsize(10pt));
real y = sqrt(3);
pair A,B,C,D,E,F,G,H;
A = (0,0);
B = (0,y);
C = (y,y);
D = (y,0);
E = ((y + 1)/2,y);
F = (y, (y - 1)/2);
G = ((y - 1)/2, 0);
H = (0,(y + 1)/2);
fill(H--B--E--cycle, gray);
draw(A--B--C--D--cycle);
draw(E--F--G--H--cycle);
[\asy]
Solution
Note that each side length is 
and 
Let the shorter side of our triangle be 
, thus the longer leg is 
. Hence, by the Pythagorean Theorem, we have ![\[(x-\sqrt{3})^2+x^2=2\]](http://latex.artofproblemsolving.com/4/3/b/43bfb185f856b6399445c3f2a3ffe675ce459426.png)
![\[2x^2-2x\sqrt{3}+1=0\]](http://latex.artofproblemsolving.com/a/3/d/a3d3d78940ddfb7a6a316dac8be206a3c8edb50d.png)
.
By the quadratic formula, we find 
. Hence, our answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}}=2-\sqrt{3}.$ (Error compiling LaTeX. Unknown error_msg)
~SirAppel
