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| ==Problem==
| | #redirect[[2023 AMC 12A Problems/Problem 9]] |
| 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?
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| <asy>
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| size(200);
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| defaultpen(linewidth(0.6pt)+fontsize(10pt));
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| real y = sqrt(3);
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| pair A,B,C,D,E,F,G,H;
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| A = (0,0);
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| B = (0,y);
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| C = (y,y);
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| D = (y,0);
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| E = ((y + 1)/2,y);
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| F = (y, (y - 1)/2);
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| G = ((y - 1)/2, 0);
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| H = (0,(y + 1)/2);
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| fill(H--B--E--cycle, gray);
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| draw(A--B--C--D--cycle);
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| draw(E--F--G--H--cycle);
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| </asy>
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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>
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| ==Solution==
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| 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>
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| <cmath>2x^2-2x\sqrt{3}+1=0</cmath>.
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| 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}=\boxed{\textbf{(C) }2-\sqrt3}.</math>
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| ~SirAppel ~ItsMeNoobieboy
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| ==Solution 2 (Area Variation of Solution 1)==
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| Looking at the diagram, knowing the square inscribed the square with area 3 is area 2. We would automatically know the area of the triangles are <math>\frac{1}{4}</math>
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| From solution 1, the base is <math>x</math> and the height <math>\sqrt{3} - x</math>. Which means <math>\frac{x(\sqrt{3} - x)}{2} = \frac{1}{4}</math>
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| We can turn this into a quadratic equation making it <math>x^2-x\sqrt{3}+\frac{1}{2} = 0</math>
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| By using the quadratic formula, we get <math>x=\frac{\sqrt{3}\pm1}{2}</math>.Therefore, the answer is <math>\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.</math>
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| ~ghfhgvghj10
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| ==See Also==
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| {{AMC10 box|year=2023|ab=A|num-b=10|num-a=12}}
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| {{MAA Notice}}
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