Difference between revisions of "2023 AMC 10A Problems/Problem 11"
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draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
draw(E--F--G--H--cycle); | draw(E--F--G--H--cycle); | ||
| − | < | + | <asy> |
<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> | ||
Revision as of 21:14, 9 November 2023
Problem
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 
~SirAppel ~ItsMeNoobieboy
See Also
| 2023 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
