Difference between revisions of "2023 AMC 12A Problems/Problem 9"
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Revision as of 20:40, 10 November 2023
- The following problem is from both the 2023 AMC 10A #11 and 2023 AMC 12A #9, so both problems redirect to this page.
Contents
Problem
A square of area is inscribed in a square of area
, creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?
Solution
Note that each side length is and
Let the shorter side of our triangle be
, thus the longer leg is
. Hence, by the Pythagorean Theorem, we have
.
By the quadratic formula, we find . Hence, our answer is
~SirAppel ~ItsMeNoobieboy
Solution 2 (Area Variation of Solution 1)
Looking at the diagram, we know that the square inscribed in the square with area has area
. Thus, the area outside of the small square is
This area is composed of
congruent triangles, so we know that each triangle has an area of
.
From solution , the base has length
and the height
, which means that
.
We can turn this into a quadratic equation: .
By using the quadratic formula, we get .Therefore, the answer is
~ghfhgvghj10 (If I made any mistakes, feel free to make minor edits)
(Clarity & formatting edits by Technodoggo)
Solution 3
Let be the ratio of the shorter leg to the longer leg, and
be the length of longer leg. The length of the shorter leg will be
.
Because the sum of two legs is the length of a side of the outside square, we have , which means
. Using the Pythagorean Theorem for the shaded right triangle, we also have
. Solving both equations, we get
. Using
to substitute
in the second equation, we get
. Hence,
. By using the quadratic formula, we get
. Because
be the ratio of the shorter leg to the longer leg, it is always less than
. Therefore, the answer is
.
~sqroot
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.