Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 4 (Pythagorean Theorem))
 
(18 intermediate revisions by 9 users not shown)
Line 5: Line 5:
 
<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>
  
−
== Solution ==  
+
== Diagram ==
−
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.
+
<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 1 ==  
 +
Note that <math>BP=BQ=DR=DS=1-s.</math> It follows that <math>\triangle BPQ</math> and <math>\triangle DRS</math> are congruent isosceles right triangles.
  
 
In <math>\triangle BPQ,</math> we have <math>PQ=BP\sqrt2,</math> or  
 
In <math>\triangle BPQ,</math> we have <math>PQ=BP\sqrt2,</math> or  
Line 17: Line 50:
 
Therefore, the answer is <cmath>s = \frac{\sqrt2}{\sqrt2 + 1}\cdot\frac{\sqrt2 - 1}{\sqrt2 - 1} = \boxed{\textbf{(C) } 2 - \sqrt{2}}.</cmath>
 
Therefore, the answer is <cmath>s = \frac{\sqrt2}{\sqrt2 + 1}\cdot\frac{\sqrt2 - 1}{\sqrt2 - 1} = \boxed{\textbf{(C) } 2 - \sqrt{2}}.</cmath>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
== Solution 2 ==
 +
Since it is an equilateral convex hexagon, all sides are the same, so we will call the side length <math>x</math>. Notice that <math>(1-x)^2+(1-x)^2 = x^2</math>. We can solve this equation which gives us our answer.
 +
<cmath>\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*}</cmath>
 +
 +
We then use the quadratic formula which gives us:
 +
 +
<cmath>\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*}</cmath>
 +
 +
Then we simplify it by dividing and crossing out 2 which gives us <math>2\pm{\sqrt2}</math> and that gives us <math>\boxed{\textbf{(C) }2-{\sqrt2}}</math>.
 +
 +
~[[OrenSH|orenbad]]
 +
 +
== Solution 3 (Area) ==
 +
 +
We can find areas in terms of <math>s.</math> From the diagram, draw in segments <math>SP</math> and <math>RQ.</math> We then have two non-shaded triangles, two shaded triangles, and a rectangle.
 +
 +
The non-shaded triangles have leg lengths of <math>1-s,</math> so they each have area <math>\frac{(1-s)^2}{2}.</math> Therefore, the total area of the two triangles is <math>(1-s)^2.</math>
 +
 +
The shaded triangles have side lengths <math>s,</math> so they each have area <math>\frac{s^2}{2}.</math> Then, we get that their combined area is <math>s^2.</math>
 +
 +
Looking at the rectangle, we find that <math>SP=RQ=s\sqrt2,</math> from 45-45-90 triangles <math>APS</math> and <math>CRQ.</math> Multiplying this with the other side length <math>s,</math> we see that the rectangle has area <math>s^2\sqrt2.</math>
 +
 +
These three expressions of area sum up to the big square, which has area 1. So, we add them up and solve:
 +
 +
<cmath>\begin{align*}
 +
(1-s)^2+s^2+s^2\sqrt2=&~1 \\
 +
s^2-2s+1+s^2+s^2\sqrt2=&~1 \\
 +
2s^2+s^2\sqrt2-2s=&~0 \\
 +
s((2+\sqrt2)s-2)=&~0 \\
 +
\end{align*}</cmath>
 +
 +
<math>s</math> cannot be 0, since it represents a positive side length. This means that <math>s</math> satisfies <math>(2+\sqrt2)s-2=0.</math> Solving, we see that
 +
 +
<cmath>s=\frac{2}{2+\sqrt2}=\frac{2}{2+\sqrt2}\cdot\frac{2-\sqrt2}{2-\sqrt2}=\frac{4-2\sqrt2}{2}=\boxed{\textbf{(C) }2-{\sqrt2}}.</cmath>
 +
 +
~UltimateDL
 +
 +
==Solution 4 (Pythagorean Theorem)==
 +
A corner of the square gives a right triangle (e.g. <math>\triangle DRS</math>) with legs <math>1-s</math> and hypotenuse <math>s</math>. It follows that
 +
 +
<cmath>\begin{align*}
 +
2(1-s)^2&=s^2 \\
 +
s^2-4s+2&=0 \\
 +
\end{align*}</cmath>
 +
 +
Completing the square gives
 +
 +
<cmath>\begin{align*}
 +
s^2-4s+4&=2 \\
 +
(s-2)^2&=2 \\
 +
s&=\pm \sqrt{2}+2 \\
 +
\end{align*}</cmath>
 +
 +
The only answer choice that appears in our solution is <math>-\sqrt{2}+2</math>, or <math>\boxed{\textbf{(C) }2-{\sqrt2}}</math>.
 +
 +
~MrThinker
  
 
==Video Solution 1 (Quick and Easy)==
 
==Video Solution 1 (Quick and Easy)==
Line 22: Line 120:
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
 +
 +
==Video Solution 2==
 +
https://youtu.be/P4Oiyf_JjHo
  
 
== See Also ==
 
== See Also ==

Latest revision as of 10:03, 15 February 2024

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 1

Note that $BP=BQ=DR=DS=1-s.$ It follows that $\triangle BPQ$ and $\triangle DRS$ are congruent 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 $(1-x)^2+(1-x)^2 = x^2$. 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

Solution 3 (Area)

We can find areas in terms of $s.$ From the diagram, draw in segments $SP$ and $RQ.$ We then have two non-shaded triangles, two shaded triangles, and a rectangle.

The non-shaded triangles have leg lengths of $1-s,$ so they each have area $\frac{(1-s)^2}{2}.$ Therefore, the total area of the two triangles is $(1-s)^2.$

The shaded triangles have side lengths $s,$ so they each have area $\frac{s^2}{2}.$ Then, we get that their combined area is $s^2.$

Looking at the rectangle, we find that $SP=RQ=s\sqrt2,$ from 45-45-90 triangles $APS$ and $CRQ.$ Multiplying this with the other side length $s,$ we see that the rectangle has area $s^2\sqrt2.$

These three expressions of area sum up to the big square, which has area 1. So, we add them up and solve:

\begin{align*} (1-s)^2+s^2+s^2\sqrt2=&~1 \\ s^2-2s+1+s^2+s^2\sqrt2=&~1 \\ 2s^2+s^2\sqrt2-2s=&~0 \\ s((2+\sqrt2)s-2)=&~0 \\ \end{align*}

$s$ cannot be 0, since it represents a positive side length. This means that $s$ satisfies $(2+\sqrt2)s-2=0.$ Solving, we see that

\[s=\frac{2}{2+\sqrt2}=\frac{2}{2+\sqrt2}\cdot\frac{2-\sqrt2}{2-\sqrt2}=\frac{4-2\sqrt2}{2}=\boxed{\textbf{(C) }2-{\sqrt2}}.\]

~UltimateDL

Solution 4 (Pythagorean Theorem)

A corner of the square gives a right triangle (e.g. $\triangle DRS$) with legs $1-s$ and hypotenuse $s$. It follows that

\begin{align*} 2(1-s)^2&=s^2 \\ s^2-4s+2&=0 \\ \end{align*}

Completing the square gives

\begin{align*} s^2-4s+4&=2 \\ (s-2)^2&=2 \\ s&=\pm \sqrt{2}+2 \\ \end{align*}

The only answer choice that appears in our solution is $-\sqrt{2}+2$, or $\boxed{\textbf{(C) }2-{\sqrt2}}$.

~MrThinker

Video Solution 1 (Quick and Easy)

https://youtu.be/uXG8xTGwx-8

~Education, the Study of Everything

Video Solution 2

https://youtu.be/P4Oiyf_JjHo

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_5&oldid=215528"