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

Revision as of 18:28, 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

@@ Line 4: / Line 4: @@
 <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>
+== 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("<math>A</math>",A,NW,linewidth(4));
+dot("<math>B</math>",B,NE,linewidth(4));
+dot("<math>C</math>",C,SE,linewidth(4));
+dot("<math>D</math>",D,SW,linewidth(4));
+dot("<math>P</math>",P,N,linewidth(4));
+dot("<math>Q</math>",Q,E,linewidth(4));
+dot("<math>R</math>",R,(0,-1),linewidth(4));
+dot("<math>S</math>",S,W,linewidth(4));
+label("<math>s</math>",midpoint(A--P),N,red);
+label("<math>s</math>",midpoint(P--Q),NE,red);
+label("<math>s</math>",midpoint(Q--C),E,red);
+label("<math>s</math>",midpoint(C--R),(0,-1),red);
+label("<math>s</math>",midpoint(R--S),SW,red);
+label("<math>s</math>",midpoint(S--A),W,red);
+[/asy]
+~MRENTHUSIASM
 == Solution ==

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