Difference between revisions of "2023 AMC 12A Problems/Problem 9"
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==Solution== | ==Solution== | ||
− | + | == Solution 1 (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>. |
− | |||
− | |||
− | + | ==Solution 2 (Area)== | |
− | |||
− | ==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\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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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>. | 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 length | + | 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 | ~sqroot | ||
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==Solution 4== | ==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 < | + | 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 | ~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 == | ||
+ | |||
+ | 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>(\sqrt{3}-x)^2+x^2=2</cmath> | ||
+ | <cmath>2x^2-2x\sqrt{3}+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> | ||
+ | |||
+ | ~SirAppel ~ItsMeNoobieboy | ||
+ | |||
+ | ==Video Solution by Power Solve (easy to digest!)== | ||
+ | https://www.youtube.com/watch?v=7KXT1pI-i64 | ||
+ | |||
+ | ==Video Solution (⚡Under 4 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== | ==Video Solution== | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/N2lyYRMuZuk?si=99Ir3nUQxTkJDiVm | ||
+ | |||
+ | ~Math-X | ||
+ | |||
==See Also== | ==See Also== |
Latest revision as of 17:43, 28 July 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
- 3 Solution 1 (Manipulation)
- 4 Solution 2 (Area)
- 5 Solution 3
- 6 Solution 4
- 7 Solution 5
- 8 Solution 6
- 9 Solution 7
- 10 Video Solution by Power Solve (easy to digest!)
- 11 Video Solution (⚡Under 4 min⚡)
- 12 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 13 Video Solution
- 14 Video Solution by Math-X (First understand the problem!!!)
- 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
Solution 1 (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
.
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
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
.
By the quadratic formula, we find . Hence, our answer is
~SirAppel ~ItsMeNoobieboy
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)
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/N2lyYRMuZuk?si=99Ir3nUQxTkJDiVm
~Math-X
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.