Difference between revisions of "2022 AMC 10A Problems/Problem 5"
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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> | ||
+ | |||
+ | == 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 == | == Solution == |
Revision as of 17:28, 16 November 2022
Problem
Square has side length . Points , , , and each lie on a side of such that is an equilateral convex hexagon with side length . What is ?
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,NW,linewidth(4)); dot("",B,NE,linewidth(4)); dot("",C,SE,linewidth(4)); dot("",D,SW,linewidth(4)); dot("",P,N,linewidth(4)); dot("",Q,E,linewidth(4)); dot("",R,(0,-1),linewidth(4)); dot("",S,W,linewidth(4)); label("",midpoint(A--P),N,red); label("",midpoint(P--Q),NE,red); label("",midpoint(Q--C),E,red); label("",midpoint(C--R),(0,-1),red); label("",midpoint(R--S),SW,red); label("",midpoint(S--A),W,red); [/asy] ~MRENTHUSIASM
Solution
Note that It follows that and are isosceles right triangles.
In we have or Therefore, the answer is ~MRENTHUSIASM
Solution 2
Since it is an equilateral convex hexagon, all sides are the same, so we will call the side length . Notice that . We can solve this equation which gives us our answer.
We then use the quadratic formula which gives us:
Then we simplify it by dividing and crossing out 2 which gives us and that gives us .
~orenbad
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
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.