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

(Solution)
(Solution)
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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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== Diagram ==
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[asy]
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/* Made by MRENTHUSIASM */
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size(200);
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real s = 2-sqrt(2);
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pair A, B, C, D, P, Q, R, S;
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A = (0,1);
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B = (1,1);
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C = (1,0);
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D = (0,0);
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P = A + (s,0);
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Q = B - (0,1-s);
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R = C - (s,0);
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S = D + (0,1-s);
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fill(A--P--Q--C--R--S--cycle,yellow);
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draw(A--B--C--D--cycle^^P--Q^^R--S);
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dot("<math>A</math>",A,NW,linewidth(4));
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dot("<math>B</math>",B,NE,linewidth(4));
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dot("<math>C</math>",C,SE,linewidth(4));
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dot("<math>D</math>",D,SW,linewidth(4));
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dot("<math>P</math>",P,N,linewidth(4));
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dot("<math>Q</math>",Q,E,linewidth(4));
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dot("<math>R</math>",R,(0,-1),linewidth(4));
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dot("<math>S</math>",S,W,linewidth(4));
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label("<math>s</math>",midpoint(A--P),N,red);
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label("<math>s</math>",midpoint(P--Q),NE,red);
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label("<math>s</math>",midpoint(Q--C),E,red);
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label("<math>s</math>",midpoint(C--R),(0,-1),red);
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label("<math>s</math>",midpoint(R--S),SW,red);
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label("<math>s</math>",midpoint(S--A),W,red);
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[/asy]
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~MRENTHUSIASM
  
 
== Solution ==  
 
== Solution ==  

Revision as of 17:28, 16 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 $s$?

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

Diagram

[asy] /* Made by MRENTHUSIASM */ size(200); real s = 2-sqrt(2); pair A, B, C, D, P, Q, R, S; A = (0,1); B = (1,1); C = (1,0); D = (0,0); P = A + (s,0); Q = B - (0,1-s); R = C - (s,0); S = D + (0,1-s); fill(A--P--Q--C--R--S--cycle,yellow); draw(A--B--C--D--cycle^^P--Q^^R--S); dot("$A$",A,NW,linewidth(4)); dot("$B$",B,NE,linewidth(4)); dot("$C$",C,SE,linewidth(4)); dot("$D$",D,SW,linewidth(4)); dot("$P$",P,N,linewidth(4)); dot("$Q$",Q,E,linewidth(4)); dot("$R$",R,(0,-1),linewidth(4)); dot("$S$",S,W,linewidth(4)); label("$s$",midpoint(A--P),N,red); label("$s$",midpoint(P--Q),NE,red); label("$s$",midpoint(Q--C),E,red); label("$s$",midpoint(C--R),(0,-1),red); label("$s$",midpoint(R--S),SW,red); label("$s$",midpoint(S--A),W,red); [/asy] ~MRENTHUSIASM

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

Solution 2

Since it is an equilateral convex hexagon, all sides are the same, so we will call the side length $x$. Notice that $(1-x)^2\cdot(1-x)^2 = x^2$. We can solve this equation which gives us our answer. \begin{align*} 1+x^2-2x+1+x^2-2x &= x^2 \\ 2x^2-4x+2 &= x^2 \\ x^2-4x+2 &= 0 \\ \end{align*}

We then use the quadratic formula which gives us:

\begin{align*} x &= \frac{4\:\pm\sqrt{4^2-4\cdot 1\cdot 2}}{2\cdot 1} \\ &= \frac{4\:\pm\sqrt{8}}{2} \\ &= \frac{4\:\pm2\sqrt{2}}{2} \\ \end{align*}

Then we simplify it by dividing and crossing out 2 which gives us $2\pm{\sqrt2}$ and that gives us $\boxed{\textbf{(C) }2-{\sqrt2}}$.

~orenbad

Video Solution 1 (Quick and Easy)

https://youtu.be/uXG8xTGwx-8

~Education, the Study of Everything

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
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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