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Difference between revisions of "2024 AMC 10A Problems/Problem 22"

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~i_am_suk_at_math(very minor latex edits)
 
~i_am_suk_at_math(very minor latex edits)
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==Solution 3==
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Let's start by looking at kite <math>\mathcal K</math>. We can quickly deduce based off of the side lengths that the kite can be split into two <math>30-60-90</math> triangles. Going back to the triangle <math>\triangle ABC</math>, focus on side <math>AB</math>. There are <math>4</math> kites, they are all either reflected over the line <math>AB</math> or a line perpendicular to <math>AB</math>, meaning the length of <math>AB</math> can be split up into 4 equal parts.
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Pick out the bottom-left kite, and we can observe that the kite and the triangle formed by the intersection of the kite and <math>\Delta ABC</math> share a <math>60</math> degree angle. (this was deduced from the <math>30-60-90</math> triangles in the kite) The line AB and the right side of the kite are perpendicular, forming a <math>90^{\circ}</math> angle. Because that is also a <math>30-60-90</math> triangle with a hypotenuse of <math>\sqrt3</math>, so we find the length of AB to be <math>4*3/2</math>, which is <math>6</math>.
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Then, we can drop an altitude from <math>C</math> to <math>AB</math>. We know that will be equivalent to the sum of the longer side of the kite and the shorter side of the triangle formed by the intersection of the kite and <math>\Delta ABC</math>. (Look at the line formed on the left of <math>C</math> that drops down to <math>AB</math> if you are confused) We already have those values from the <math>30-60-90</math> triangles, so we can just plug it into the triangle area formula, <math>bh/2</math>. We get <cmath>6\cdot\dfrac{\sqrt3+\frac{\sqrt3}{2}}{2}\rightarrow3\cdot(\sqrt3+\dfrac{\sqrt3}{2})\rightarrow3\cdot\dfrac{\sqrt3}{2}\rightarrow\boxed{\textbf{(B) }\dfrac92\sqrt3}</cmath>
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~YTH (Need help with Latex and formatting)
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~WIP (Header)
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~Tacos_are_yummy_1 (<math>\LaTeX</math> & Formatting)
  
 
== Video Solution by Innovative Minds ==
 
== Video Solution by Innovative Minds ==

Revision as of 16:10, 9 November 2024

Problem

Let $\mathcal K$ be the kite formed by joining two right triangles with legs $1$ and $\sqrt3$ along a common hypotenuse. Eight copies of $\mathcal K$ are used to form the polygon shown below. What is the area of triangle $\Delta ABC$?

Asset-ddfea426a1acee64ea44467d8aa8797a.png

$\textbf{(A) }2+3\sqrt3\qquad\textbf{(B) }\dfrac92\sqrt3\qquad\textbf{(C) }\dfrac{10+8\sqrt3}{3}\qquad\textbf{(D) }8\qquad\textbf{(E) }5\sqrt3$

Solution 3

202410A 23.png (latexing a WIP) ~mathboy282 ~Thesmartgreekmathdude

Solution 4

Let the point of intersection of $AB$ and the kite with $A$ as vertex be $D$.

Let the left kite with $C$ as a vertex touch the kite with $A$ as vertex at point $E$.

$\triangle ADE$ is a $30-60-90$ so $AD = \frac{3}{2}$ and $DE = \frac{\sqrt3}{2}$.

So, $AB = 4\cdot AD = 6$ and $CD=CE+DE= \frac{3\sqrt3}{2}$, and the area is $\frac12\cdot AB \cdot CD = \boxed{ \textbf{(B) } \frac{9}{2} \sqrt3}.$

~Mintylemon66

Solution 1

First, we should find the length of $AB$. In order to do this, as we see in the diagram, it can be split into 4 sections. Since diagram $K$ shows us that it is made up of two ${30,60,90}$ triangles, then the triangle outlined in red must be a $30,60,90$ triangle. Also, since we know the length of the longest side is ${\sqrt3}$, then the side we are looking for, which is outlined in blue, must be $\frac{3}{2}$ by the ${1,\sqrt3, 2}$ relationship of ${30,60,90}$ triangles. Therefore the long side that is the base of the triangle we are looking for must be {6}.


Now all we have to do is find the height. We can split the height into 2 sections, the green and the light green. The green section must be ${\sqrt3}$, as $K$ shows us. Also, the light green section must be equal to ${\frac{\sqrt3}{2}}$, as in the previous paragraph, the triangle outlined in red is $30,60,90$. Then, the green section, which is the height, must be ${\sqrt3}+{\frac{\sqrt3}{2}}$, which is just ${\frac{3\sqrt3}{2}}$.


Then the area of the triangle must be ${\frac{1}{2}}*base*height$, which is just $\boxed{ \textbf{(B) } \frac{9}{2} \sqrt3}.$

~Solution by HappySharks

Solution 2

Let $\mathcal K$ be quadrilateral $MNOP$. Drawing line $MO$ splits the triangle into $\Delta MNO$. Drawing the altitude from $N$ to point $Q$ on line $MO$, we know $NQ$ is $\frac{\sqrt{3}}{2}$, $MQ$ is $\frac{3}{2}$, and $QO$ is $\frac{1}{2}$.

Screenshot 2024-11-08 2.33.52 PM.png

Due to the many similarities present, we can find that $AB$ is $4(MQ)$, and the height of $\Delta ABC$ is $NQ+MN$

$AB$ is $4(\frac{3}{2})=6$ and the height of $\Delta ABC$ is $\sqrt3+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$.

Solving for the area of $\Delta ABC$ gives $6\cdot\frac{3\sqrt{3}}{2}\cdot\frac{1}{2}$ which is $\boxed{\textbf{(B) }\dfrac92\sqrt3}$

~9897 (latex beginner here)

~i_am_suk_at_math(very minor latex edits)

Solution 3

Let's start by looking at kite $\mathcal K$. We can quickly deduce based off of the side lengths that the kite can be split into two $30-60-90$ triangles. Going back to the triangle $\triangle ABC$, focus on side $AB$. There are $4$ kites, they are all either reflected over the line $AB$ or a line perpendicular to $AB$, meaning the length of $AB$ can be split up into 4 equal parts.

Pick out the bottom-left kite, and we can observe that the kite and the triangle formed by the intersection of the kite and $\Delta ABC$ share a $60$ degree angle. (this was deduced from the $30-60-90$ triangles in the kite) The line AB and the right side of the kite are perpendicular, forming a $90^{\circ}$ angle. Because that is also a $30-60-90$ triangle with a hypotenuse of $\sqrt3$, so we find the length of AB to be $4*3/2$, which is $6$.

Then, we can drop an altitude from $C$ to $AB$. We know that will be equivalent to the sum of the longer side of the kite and the shorter side of the triangle formed by the intersection of the kite and $\Delta ABC$. (Look at the line formed on the left of $C$ that drops down to $AB$ if you are confused) We already have those values from the $30-60-90$ triangles, so we can just plug it into the triangle area formula, $bh/2$. We get \[6\cdot\dfrac{\sqrt3+\frac{\sqrt3}{2}}{2}\rightarrow3\cdot(\sqrt3+\dfrac{\sqrt3}{2})\rightarrow3\cdot\dfrac{\sqrt3}{2}\rightarrow\boxed{\textbf{(B) }\dfrac92\sqrt3}\]

~YTH (Need help with Latex and formatting)

~WIP (Header)

~Tacos_are_yummy_1 ($\LaTeX$ & Formatting)

Video Solution by Innovative Minds

https://www.youtube.com/watch?v=bhC58BB3kJA

~i_am_suk_at_math_2


See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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

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