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Difference between revisions of "2024 AMC 10A Problems/Problem 22"
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==Solution 2==
 
==Solution 2==
  
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~YTH (Working on it right now, please don't interfere. Thanks :) also if someone could help me create the solution 2 header that would be awesome)
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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 <math>2</math> <math>30</math>-<math>60</math>-<math>90</math> triangles. Going back to the triangle ABC, focus on side AB. There are <math>4</math> 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.
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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</math>-<math>60</math>-<math>90</math> triangles in the kite) The line AB and the right side of the kite are perpendicular, forming a <math>90</math> degree angle. Because that is also a <math>30</math>-<math>60</math>-<math>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 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 <math>\Delta ABC</math>. (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 <math>30</math>-<math>60</math>-<math>90</math> triangles, so we can just plug it into the triangle area formula, <math>bh/2</math>. We get <math>6*(\sqrt3+\sqrt3/2)/2</math> ==> <math>3*(\sqrt3+\sqrt3/2)</math> ==> <math>3*\sqrt3/2</math> ==> <math>\textbf{(B) }\dfrac92\sqrt3\qquad</math>
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~YTH (Need help with Latex and formatting)
  
 
~WIP (Header)
 
~WIP (Header)
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==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
 
{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:37, 8 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 1

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 $\sqrt3/2$, MQ is $3/2$, and QO is $1/2$.

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 $4(3/2)=6$ and the height of $\Delta ABC$ is $\sqrt3+\sqrt3/2=3\sqrt3/2$.

Solving for the area of $\Delta ABC$ gives $6*3\sqrt3/2*1/2$ which is $\textbf{(B) }\dfrac92\sqrt3\qquad$

~9897 (latex beginner here)

Solution 2

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 $2$ $30$-$60$-$90$ triangles. Going back to the 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 $90$ degree 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 $6*(\sqrt3+\sqrt3/2)/2$ ==> $3*(\sqrt3+\sqrt3/2)$ ==> $3*\sqrt3/2$ ==> $\textbf{(B) }\dfrac92\sqrt3\qquad$

~YTH (Need help with Latex and formatting)

~WIP (Header)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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