2022 AMC 10A Problems/Problem 5
Contents
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
~MRENTHUSIASM
Solution 1
Note that It follows that and are congruent 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 .
Solution 3 (Area)
We can find areas in terms of From the diagram, draw in segments and We then have two non-shaded triangles, two shaded triangles, and a rectangle.
The non-shaded triangles have leg lengths of so they each have area Therefore, the total area of the two triangles is
The shaded triangles have side lengths so they each have area Then, we get that their combined area is
Looking at the rectangle, we find that from 45-45-90 triangles and Multiplying this with the other side length we see that the rectangle has area
These three expressions of area sum up to the big square, which has area 1. So, we add them up and solve:
cannot be 0, since it represents a positive side length. This means that satisfies Solving, we see that
~UltimateDL
Solution 4 (Pythagorean Theorem)
A corner of the square gives a right triangle (e.g. ) with legs and hypotenuse . It follows that
Completing the square gives
The only answer choice that appears in our solution is , or .
~MrThinker
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 | |
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All AMC 10 Problems and Solutions |
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