Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12A Problems/Problem 9"

(→‎Solution)
m (→‎Video Solution (⚡Under 4 min⚡))
 
(44 intermediate revisions by 19 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2023 AMC 10A Problems/Problem 11|2023 AMC 10A #11]] and [[2023 AMC 12A Problems/Problem 9|2023 AMC 12A #9]]}}
 +
 
==Problem==
 
==Problem==
 
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?
 
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?
Line 21: Line 23:
 
<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>.  
−
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>
+
Hence, by the Pythagorean Theorem, we have <cmath>(\sqrt{3}-x)^2+x^2=(\sqrt{2})^2</cmath>
−
<cmath>2x^2-2x\sqrt{3}+1=0</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 <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>
+
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>
  
−
~SirAppel ~ItsMeNoobieboy
+
~semisteve
 +
~SirAppel  
 +
~ItsMeNoobieboy
  
−
==Solution 2 (Area Variation of Solution 1)==
+
==Solution 2 (Area)==
  
 
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(\sqrt{3} - x)}{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>.  
Line 43: Line 48:
  
 
(Clarity & formatting edits by Technodoggo)
 
(Clarity & formatting edits by Technodoggo)
 +
 +
==Solution 3==
 +
 +
Let <math>x</math> be the ratio of the shorter leg to the longer leg, and <math>y</math> be the length of longer leg. The length of the shorter leg will be <math>xy</math>.
 +
 +
Because the sum of two legs is the side length of the outside square, we have <math>xy + y = \sqrt{3}</math>, which means <math>(xy)^2 + y^2 + 2xy^2 = 3</math>. Using the Pythagorean Theorem for the shaded right triangle, we also have <math>(xy)^2 + y^2 = 2</math>. Solving both equations, we get <math>2xy^2 = 1</math>. Using <math>y^2=\frac{1}{2x}</math> to substitute <math>y</math> in the second equation, we get <math>x^2\cdot \frac{1}{2x} + \frac{1}{2x} = 2</math>. Hence, <math>x^2 - 4x + 1 = 0</math>. By using the quadratic formula, we get <math>x=2\pm \sqrt3</math>. Because <math>x</math> be the ratio of the shorter leg to the longer leg, it is always less than <math>1</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }2-\sqrt3}</math>.
 +
 +
~sqroot
 +
 +
==Solution 4==
 +
 +
The side length of the bigger square is equal to <math>\sqrt{3}</math>, while the side length of the smaller square is <math>\sqrt{2}</math>. Let <math>x</math> be the shorter leg and <math>y</math> be the longer one. Clearly, <math>x+y=\sqrt{3}</math>, and <math>xy=\frac{1}{2}</math>. Using Vieta's to build a quadratic, we get <cmath>x^2-\sqrt{3}x+\frac{1}{2}=0.</cmath> Solving, we get <math>x=\frac{\sqrt{3}-1}{2}</math> and <math>y=\frac{\sqrt{3}+1}{2}</math>. Thus,  we find <math>\frac{x}{y}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)\cdot(\sqrt{3}-1)}=\frac{4-2\sqrt{3}}{2}=\boxed{\textbf{(C) }2-\sqrt3}</math>.
 +
 +
~vadava_lx
 +
 +
 +
==Solution 5==
 +
 +
Let <math>\theta</math> be the angle opposite the smaller leg. We want to find <math>\tan\theta</math>.
 +
 +
The area of the triangle is <math>\frac{1}{2}\left(\sqrt{2}\sin\theta\right)\left(\sqrt{2}\cos\theta\right)=\frac{1}{2}\sin 2\theta=\frac{1}{4},</math> which implies <math>\sin 2\theta = \frac{1}{2},</math> or <math>\theta=15^\circ</math>. Therefore <math>\tan \theta = \boxed{\textbf{(C) }2-\sqrt3}</math>
 +
 +
==Solution 6==
 +
 +
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 (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!)==
 +
https://www.youtube.com/watch?v=7KXT1pI-i64
 +
 +
==Video Solution (⚡Under 5 min⚡)==
 +
https://youtu.be/OiBUPU1bULc
 +
 +
~Education, the Study of Everything
 +
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 +
https://www.youtube.com/watch?v=IVgzVS86Ogo
 +
 +
==Video Solution==
 +
 +
https://youtu.be/jL2k7MFgfzQ
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
  
 
==See Also==
 
==See Also==

Latest revision as of 00:08, 11 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.

Problem

A square of area $2$ is inscribed in a square of area $3$, 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]

$\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1$

Solution 1

The side lengths of the inner square and outer square are $\sqrt{2}$ and $\sqrt{3}$ respectively. Let the shorter side of our triangle be $x$, thus the longer leg is $\sqrt{3}-x$. Hence, by the Pythagorean Theorem, we have \[(\sqrt{3}-x)^2+x^2=(\sqrt{2})^2\] \[3-2\sqrt{3}x+x^2+x^2=2\] \[2x^2-2\sqrt{3}x+1=0\]

By the quadratic formula, we find that $x=\frac{\sqrt{3}\pm1}{2}$, so the answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.$

~semisteve ~SirAppel ~ItsMeNoobieboy

Solution 2 (Area)

Looking at the diagram, we know that the square inscribed in the square with area $3$ has area $2$. Thus, the area outside of the small square is $3-2=1.$ This area is composed of $4$ congruent triangles, so we know that each triangle has an area of $\dfrac14$.

From solution $1$, the base (short side of the triangle) has a length $x$ and the height $\sqrt{3} - x$, which means that $\frac{x\left(\sqrt{3} - x\right)}{2} = \frac{1}{4}$.

We can turn this into a quadratic equation: $x^2-x\sqrt{3}+\frac{1}{2} = 0$.

By using the quadratic formula, we get $x=\frac{\sqrt{3}\pm1}{2}$.Therefore, the answer is $\frac{\sqrt{3}-1}{\sqrt{3}+1}=\boxed{\textbf{(C) }2-\sqrt3}.$

~ghfhgvghj10 (If I made any mistakes, feel free to make minor edits)

(Clarity & formatting edits by Technodoggo)

Solution 3

Let $x$ be the ratio of the shorter leg to the longer leg, and $y$ be the length of longer leg. The length of the shorter leg will be $xy$.

Because the sum of two legs is the side length of the outside square, we have $xy + y = \sqrt{3}$, which means $(xy)^2 + y^2 + 2xy^2 = 3$. Using the Pythagorean Theorem for the shaded right triangle, we also have $(xy)^2 + y^2 = 2$. Solving both equations, we get $2xy^2 = 1$. Using $y^2=\frac{1}{2x}$ to substitute $y$ in the second equation, we get $x^2\cdot \frac{1}{2x} + \frac{1}{2x} = 2$. Hence, $x^2 - 4x + 1 = 0$. By using the quadratic formula, we get $x=2\pm \sqrt3$. Because $x$ be the ratio of the shorter leg to the longer leg, it is always less than $1$. Therefore, the answer is $\boxed{\textbf{(C) }2-\sqrt3}$.

~sqroot

Solution 4

The side length of the bigger square is equal to $\sqrt{3}$, while the side length of the smaller square is $\sqrt{2}$. Let $x$ be the shorter leg and $y$ be the longer one. Clearly, $x+y=\sqrt{3}$, and $xy=\frac{1}{2}$. Using Vieta's to build a quadratic, we get \[x^2-\sqrt{3}x+\frac{1}{2}=0.\] Solving, we get $x=\frac{\sqrt{3}-1}{2}$ and $y=\frac{\sqrt{3}+1}{2}$. Thus, we find $\frac{x}{y}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)\cdot(\sqrt{3}-1)}=\frac{4-2\sqrt{3}}{2}=\boxed{\textbf{(C) }2-\sqrt3}$.

~vadava_lx


Solution 5

Let $\theta$ be the angle opposite the smaller leg. We want to find $\tan\theta$.

The area of the triangle is $\frac{1}{2}\left(\sqrt{2}\sin\theta\right)\left(\sqrt{2}\cos\theta\right)=\frac{1}{2}\sin 2\theta=\frac{1}{4},$ which implies $\sin 2\theta = \frac{1}{2},$ or $\theta=15^\circ$. Therefore $\tan \theta = \boxed{\textbf{(C) }2-\sqrt3}$

Solution 6

Allow a and b to be the sides of a triangle. WLOG, suppose $a > b.$We want to find $\frac{b}{a}$. Notice that the area of a triangle is $\frac{3-2}{4}$, which results in $\frac{1}{4}$. Thus, $ab = \frac{1}{2}$. However, the square of the hypotenuse of this triangle is $a^2+b^2$, but also $2$. We can write $b$ as $\frac{1}{2a}$, and then plug it in. We get $a^2+\frac{1}{4a^2} = 2$, so $4a^4-8a^2+1 = 0$. Applying the quadratic formula, $a^2 = \frac{8 \pm 4\sqrt{3}}{2}$, or $4 \pm 2\sqrt3$. However, since $a$ and $b$ must both be solutions of the quadratic, since both equations were cyclic. Since $a>b$, then $a^2 = 4+2\sqrt3$, and $b^2 = 4-2\sqrt3$. To find $\frac{b}{a}$, we can simply find the square root of $\frac{b^2}{a^2}$. This is $\sqrt{\frac{4-2\sqrt3}{4+2\sqrt3}} = \sqrt{\frac{2-\sqrt3}{2+\sqrt3}} = \sqrt{(2-\sqrt3)(2-\sqrt3)} = \boxed {2-\sqrt3}$, so the answer is $\boxed {C}$. - Sepehr2010

Solution 7 (Manipulation)

Let $a$ be the length of the shorter leg and $b$ be the longer leg. By the Pythagorean theorem, we can derive that $a^2+b^2=2$. Using area we can also derive that $2ab=1$. $(a+b)^2=3$ as given in the diagram, we can find that $(a-b)^2=1$ because $4ab=2$. This means that $b+a=\sqrt{3}$ and $b-a=1$. Adding the equations gives $b=\frac{\sqrt{3}+1}{2}$ and when $b$ is plugged in $a = \frac{\sqrt{3}-1}{2}$. Rationalizing the denominators gives us $\boxed{\textbf{(C) } 2-\sqrt{3}}$.

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 5 min⚡)

https://youtu.be/OiBUPU1bULc

~Education, the Study of Everything

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=IVgzVS86Ogo

Video Solution

https://youtu.be/jL2k7MFgfzQ

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_9&oldid=233191"