Difference between revisions of "2022 AMC 10A Problems/Problem 5"
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== Solution 1 == | == 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 isosceles right triangles. | + | 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 |
Revision as of 07:11, 24 September 2023
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 | |
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.