Difference between revisions of "2023 AMC 12A Problems/Problem 9"
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==Solution 2 (Area Variation of Solution 1)== | ==Solution 2 (Area Variation of Solution 1)== | ||
| − | Looking at the diagram, | + | Looking at the diagram, we know that the square inscribed in the square with area <math>3</math> has area <math>2</math>. Thus, the area outside of the small square is <math>3-2=1.</math> This area is composed of <math>4</math> congruent triangles, so we know that each triangle has an area of <math>\dfrac14</math>. |
| − | From solution 1, the base | + | From solution <math>1</math>, the base has length <math>x</math> and the height <math>\sqrt{3} - x</math>, which means that <math>\frac{x(\sqrt{3} - x)}{2} = \frac{1}{4}</math>. |
| − | We can turn this into a quadratic equation | + | We can turn this into a quadratic equation: <math>x^2-x\sqrt{3}+\frac{1}{2} = 0</math>. |
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> | 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> | ||
~ghfhgvghj10 (If I made any mistakes, feel free to make minor edits) | ~ghfhgvghj10 (If I made any mistakes, feel free to make minor edits) | ||
| + | |||
| + | (Clarity & formatting edits by Technodoggo) | ||
==See Also== | ==See Also== | ||
Revision as of 23:22, 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]](http://latex.artofproblemsolving.com/7/6/3/763eea423cc19d0e0a602d3e907bd5edc03cc6b7.png)
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
Solution 2 (Area Variation of Solution 1)
Looking at the diagram, we know that the square inscribed in the square with area has area . Thus, the area outside of the small square is 
This area is composed of 
congruent triangles, so we know that each triangle has an area of 
.
From solution 
, the base has length and the height 
, which means that 
.
We can turn this into a quadratic equation: 
.
By using the quadratic formula, we get .Therefore, the answer is
~ghfhgvghj10 (If I made any mistakes, feel free to make minor edits)
(Clarity & formatting edits by Technodoggo)
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 | ||
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
