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

(→‎Solution 3)
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==Solution 1==
 
==Solution 1==
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Let <math>\mathcal K</math> be quadrilateral MNOP. Drawing line MO splits the triangle into <math>\Delta MNO</math>.
+
Let <math>\mathcal K</math> be quadrilateral <math>MNOP</math>. Drawing line <math>MO</math> splits the triangle into <math>\Delta MNO</math>.
−
Drawing the altitude from N to point Q on line MO, we know NQ is <math>\sqrt3/2</math>, MQ is <math>3/2</math>, and QO is <math>1/2</math>.
+
Drawing the altitude from <math>N</math> to point <math>Q</math> on line <math>MO</math>, we know <math>NQ</math> is <math>\frac{\sqrt{3}}{2}</math>, <math>MQ</math> is <math>\frac{3}{2}</math>, and <math>QO</math> is <math>\frac{1}{2}</math>.
  
 
[[File:Screenshot 2024-11-08 2.33.52 PM.png]]
 
[[File:Screenshot 2024-11-08 2.33.52 PM.png]]
  
−
Due to the many similarities present, we can find that AB is <math>4(MQ)</math>, and the height of <math>\Delta ABC</math> is <math>NQ+MN</math>
+
Due to the many similarities present, we can find that <math>AB</math> is <math>4(MQ)</math>, and the height of <math>\Delta ABC</math> is <math>NQ+MN</math>
  
−
AB is <math>4(3/2)=6</math> and the height of <math>\Delta ABC</math> is <math>\sqrt3+\sqrt3/2=3\sqrt3/2</math>.
+
<math>AB</math> is <math>4(\frac{3}{2})=6</math> and the height of <math>\Delta ABC</math> is <math>\sqrt3+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}</math>.
  
−
Solving for the area of <math>\Delta ABC</math> gives <math>6*3\sqrt3/2*1/2</math> which is <math>\textbf{(B) }\dfrac92\sqrt3\qquad</math>
+
Solving for the area of <math>\Delta ABC</math> gives <math>6*\frac{3\sqrt{3}}{2}*\frac{1}{2}</math> which is <math>\boxed{\textbf{(B) }\dfrac92\sqrt3}</math>
  
 
~9897 (latex beginner here)
 
~9897 (latex beginner here)
 +
~i_am_suk_at_math(very minor latex edits)
  
 
==Solution 2==
 
==Solution 2==

Revision as of 21:29, 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 $\frac{\sqrt{3}}{2}$, $MQ$ 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 $6*\frac{3\sqrt{3}}{2}*\frac{1}{2}$ which is $\boxed{\textbf{(B) }\dfrac92\sqrt3}$

~9897 (latex beginner here) ~i_am_suk_at_math(very minor latex edits)

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 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)

Solution 3

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

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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