Difference between revisions of "2023 AMC 12A Problems/Problem 9"
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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== | + | == Solution 1== |
+ | The side lengths of the inner square and outer square are <math>\sqrt{2}</math> and <math>\sqrt{3}</math> respectively. 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>(\sqrt{3}-x)^2+x^2=(\sqrt{2})^2</cmath> | ||
+ | <cmath>3-2\sqrt{3}x+x^2+x^2=2</cmath> | ||
+ | <cmath>2x^2-2\sqrt{3}x+1=0</cmath> | ||
− | + | By the quadratic formula, we find that <math>x=\frac{\sqrt{3}\pm1}{2}</math>, so the answer is <math>\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.</math> | |
− | + | ||
+ | ~semisteve | ||
+ | ~SirAppel | ||
+ | ~ItsMeNoobieboy | ||
==Solution 2 (Area)== | ==Solution 2 (Area)== | ||
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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>. | 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 <math>1</math>, the base has length <math>x</math> and the height <math>\sqrt{3} - x</math>, which means that <math>\frac{x\left(\sqrt{3} - x\right)}{2} = \frac{1}{4}</math>. | + | From solution <math>1</math>, the base (short side of the triangle) has a length <math>x</math> and the height <math>\sqrt{3} - x</math>, which means that <math>\frac{x\left(\sqrt{3} - x\right)}{2} = \frac{1}{4}</math>. |
We can turn this into a quadratic equation: <math>x^2-x\sqrt{3}+\frac{1}{2} = 0</math>. | We can turn this into a quadratic equation: <math>x^2-x\sqrt{3}+\frac{1}{2} = 0</math>. | ||
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Allow a and b to be the sides of a triangle. WLOG, suppose <math>a > b. </math>We want to find <math>\frac{b}{a}</math>. Notice that the area of a triangle is <math>\frac{3-2}{4}</math>, which results in <math>\frac{1}{4}</math>. Thus, <math>ab = \frac{1}{2}</math>. However, the square of the hypotenuse of this triangle is <math>a^2+b^2</math>, but also <math>2</math>. We can write <math>b</math> as <math>\frac{1}{2a}</math>, and then plug it in. We get <math>a^2+\frac{1}{4a^2} = 2</math>, so <math>4a^4-8a^2+1 = 0</math>. Applying the quadratic formula, <math>a^2 = \frac{8 \pm 4\sqrt{3}}{2}</math>, or <math>4 \pm 2\sqrt3</math>. However, since <math>a</math> and <math>b</math> must both be solutions of the quadratic, since both equations were cyclic. Since <math>a>b</math>, then <math>a^2 = 4+2\sqrt3</math>, and <math>b^2 = 4-2\sqrt3</math>. To find <math>\frac{b}{a}</math>, we can simply find the square root of <math>\frac{b^2}{a^2}</math>. This is <math>\sqrt{\frac{4-2\sqrt3}{4+2\sqrt3}} = \sqrt{\frac{2-\sqrt3}{2+\sqrt3}} = \sqrt{(2-\sqrt3)(2-\sqrt3)} = \boxed {2-\sqrt3}</math>, so the answer is <math>\boxed {C}</math>. - Sepehr2010 | Allow a and b to be the sides of a triangle. WLOG, suppose <math>a > b. </math>We want to find <math>\frac{b}{a}</math>. Notice that the area of a triangle is <math>\frac{3-2}{4}</math>, which results in <math>\frac{1}{4}</math>. Thus, <math>ab = \frac{1}{2}</math>. However, the square of the hypotenuse of this triangle is <math>a^2+b^2</math>, but also <math>2</math>. We can write <math>b</math> as <math>\frac{1}{2a}</math>, and then plug it in. We get <math>a^2+\frac{1}{4a^2} = 2</math>, so <math>4a^4-8a^2+1 = 0</math>. Applying the quadratic formula, <math>a^2 = \frac{8 \pm 4\sqrt{3}}{2}</math>, or <math>4 \pm 2\sqrt3</math>. However, since <math>a</math> and <math>b</math> must both be solutions of the quadratic, since both equations were cyclic. Since <math>a>b</math>, then <math>a^2 = 4+2\sqrt3</math>, and <math>b^2 = 4-2\sqrt3</math>. To find <math>\frac{b}{a}</math>, we can simply find the square root of <math>\frac{b^2}{a^2}</math>. This is <math>\sqrt{\frac{4-2\sqrt3}{4+2\sqrt3}} = \sqrt{\frac{2-\sqrt3}{2+\sqrt3}} = \sqrt{(2-\sqrt3)(2-\sqrt3)} = \boxed {2-\sqrt3}</math>, so the answer is <math>\boxed {C}</math>. - Sepehr2010 | ||
− | == Solution 7 == | + | == Solution 7 (Manipulation)== |
− | + | Let <math>a</math> be the length of the shorter leg and <math>b</math> be the longer leg. By the Pythagorean theorem, we can derive that <math>a^2+b^2=2</math>. Using area we can also derive that <math>2ab=1</math>. <math>(a+b)^2=3</math> as given in the diagram, we can find that <math>(a-b)^2=1</math> because <math>4ab=2</math>. This means that <math>b+a=\sqrt{3}</math> and <math>b-a=1</math>. Adding the equations gives <math>b=\frac{\sqrt{3}+1}{2}</math> and when <math>b</math> is plugged in <math>a = \frac{\sqrt{3}-1}{2}</math>. Rationalizing the denominators gives us <math>\boxed{\textbf{(C) } 2-\sqrt{3}}</math>. | |
− | < | ||
− | + | ==Video Solution by Little Fermat== | |
+ | https://youtu.be/h2Pf2hvF1wE?si=jPmOlhD8haZA8-hK&t=2309 | ||
+ | ~little-fermat | ||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/GP-DYudh5qU?si=H7rtzYDnG-hbDpIQ&t=2706 | ||
− | ~ | + | ~Math-X |
==Video Solution by Power Solve (easy to digest!)== | ==Video Solution by Power Solve (easy to digest!)== | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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Latest revision as of 06:28, 5 November 2024
- The following problem is from both the 2023 AMC 10A #11 and 2023 AMC 12A #9, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Area)
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6
- 8 Solution 7 (Manipulation)
- 9 Video Solution by Little Fermat
- 10 Video Solution by Math-X (First understand the problem!!!)
- 11 Video Solution by Power Solve (easy to digest!)
- 12 Video Solution (⚡Under 4 min⚡)
- 13 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 14 Video Solution
- 15 See Also
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?
Solution 1
The side lengths of the inner square and outer square are and respectively. Let the shorter side of our triangle be , thus the longer leg is . Hence, by the Pythagorean Theorem, we have
By the quadratic formula, we find that , so the answer is
~semisteve ~SirAppel ~ItsMeNoobieboy
Solution 2 (Area)
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 (short side of the triangle) has a 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)
Solution 3
Let be the ratio of the shorter leg to the longer leg, and be the length of longer leg. The length of the shorter leg will be .
Because the sum of two legs is the side length of the outside square, we have , which means . Using the Pythagorean Theorem for the shaded right triangle, we also have . Solving both equations, we get . Using to substitute in the second equation, we get . Hence, . By using the quadratic formula, we get . Because be the ratio of the shorter leg to the longer leg, it is always less than . Therefore, the answer is .
~sqroot
Solution 4
The side length of the bigger square is equal to , while the side length of the smaller square is . Let be the shorter leg and be the longer one. Clearly, , and . Using Vieta's to build a quadratic, we get Solving, we get and . Thus, we find .
~vadava_lx
Solution 5
Let be the angle opposite the smaller leg. We want to find .
The area of the triangle is which implies or . Therefore
Solution 6
Allow a and b to be the sides of a triangle. WLOG, suppose We want to find . Notice that the area of a triangle is , which results in . Thus, . However, the square of the hypotenuse of this triangle is , but also . We can write as , and then plug it in. We get , so . Applying the quadratic formula, , or . However, since and must both be solutions of the quadratic, since both equations were cyclic. Since , then , and . To find , we can simply find the square root of . This is , so the answer is . - Sepehr2010
Solution 7 (Manipulation)
Let be the length of the shorter leg and be the longer leg. By the Pythagorean theorem, we can derive that . Using area we can also derive that . as given in the diagram, we can find that because . This means that and . Adding the equations gives and when is plugged in . Rationalizing the denominators gives us .
Video Solution by Little Fermat
https://youtu.be/h2Pf2hvF1wE?si=jPmOlhD8haZA8-hK&t=2309 ~little-fermat
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/GP-DYudh5qU?si=H7rtzYDnG-hbDpIQ&t=2706
~Math-X
Video Solution by Power Solve (easy to digest!)
https://www.youtube.com/watch?v=7KXT1pI-i64
Video Solution (⚡Under 4 min⚡)
~Education, the Study of Everything
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=IVgzVS86Ogo
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.