|
|
(10 intermediate revisions by 5 users not shown) |
Line 1: |
Line 1: |
− | A square of area <math>2</math> is inscribed in a square of area <math>3</math>, creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?
| + | #redirect[[2023 AMC 12A Problems/Problem 9]] |
− | [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] | |
− | <math>\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1</math>
| |