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Difference between revisions of "2022 AMC 10A Problems/Problem 5"

(Created page with "==Problem 5== Square <math>ABCD</math> has side length <math>1</math>. Points <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> each lie on a side of <math>A...")
 
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==Problem 5==
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==Problem==
  
 
Square <math>ABCD</math> has side length <math>1</math>. Points <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> each lie on a side of <math>ABCD</math> such that <math>APQCRS</math> is an equilateral convex hexagon with side length <math>s</math>. What is <math>s</math>?
 
Square <math>ABCD</math> has side length <math>1</math>. Points <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> each lie on a side of <math>ABCD</math> such that <math>APQCRS</math> is an equilateral convex hexagon with side length <math>s</math>. What is <math>s</math>?
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<math>\textbf{(A) } \frac{\sqrt{2}}{3} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } 2 - \sqrt{2} \qquad \textbf{(D) } 1 - \frac{\sqrt{2}}{4} \qquad \textbf{(E) } \frac{2}{3}</math>
 
<math>\textbf{(A) } \frac{\sqrt{2}}{3} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } 2 - \sqrt{2} \qquad \textbf{(D) } 1 - \frac{\sqrt{2}}{4} \qquad \textbf{(E) } \frac{2}{3}</math>
  
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[[2022 AMC 10A Problems/Problem 5|Solution]]
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== Solution ==
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Note that <math>BP=BQ=DR=DS=1-s.</math> It follows that <math>\triangle BPQ</math> and <math>\triangle DRS</math> are isosceles right triangles.
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In <math>\triangle BPQ,</math> we have <math>PQ=BP\sqrt2,</math> or
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<cmath>\begin{align*}
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s &= (1-s)\sqrt2 \\
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s &= \sqrt2 - s\sqrt2 \\
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\left(\sqrt2+1\right)s &= \sqrt2 \\
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s &= \frac{\sqrt2}{\sqrt2 + 1}.
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\end{align*}</cmath>
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Therefore, the answer is <cmath>s = \frac{\sqrt2}{\sqrt2 + 1}\cdot\frac{\sqrt2 - 1}{\sqrt2 - 1} = \boxed{\textbf{(C) } 2 - \sqrt{2}}.</cmath>
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~MRENTHUSIASM
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== See Also ==
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{{AMC10 box|year=2022|ab=A|num-b=4|num-a=6}}
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{{MAA Notice}}

Revision as of 22:31, 11 November 2022

Problem

Square $ABCD$ has side length $1$. Points $P$, $Q$, $R$, and $S$ each lie on a side of $ABCD$ such that $APQCRS$ is an equilateral convex hexagon with side length $s$. What is $s$?

$\textbf{(A) } \frac{\sqrt{2}}{3} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } 2 - \sqrt{2} \qquad \textbf{(D) } 1 - \frac{\sqrt{2}}{4} \qquad \textbf{(E) } \frac{2}{3}$

Solution

Note that $BP=BQ=DR=DS=1-s.$ It follows that $\triangle BPQ$ and $\triangle DRS$ are isosceles right triangles.

In $\triangle BPQ,$ we have $PQ=BP\sqrt2,$ or \begin{align*} s &= (1-s)\sqrt2 \\ s &= \sqrt2 - s\sqrt2 \\ \left(\sqrt2+1\right)s &= \sqrt2 \\ s &= \frac{\sqrt2}{\sqrt2 + 1}. \end{align*} Therefore, the answer is \[s = \frac{\sqrt2}{\sqrt2 + 1}\cdot\frac{\sqrt2 - 1}{\sqrt2 - 1} = \boxed{\textbf{(C) } 2 - \sqrt{2}}.\] ~MRENTHUSIASM

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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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