2022 AMC 10A Problems/Problem 5
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 |
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