2023 AMC 12A Problems/Problem 9
- The following problem is from both the 2023 AMC 10A #11 and 2023 AMC 12A #9, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution
- 3 Solution 1 (Manipulation)
- 4 Solution 2 (Area)
- 5 Solution 3
- 6 Solution 4
- 7 Solution 5
- 8 Solution 6
- 9 Solution 6
- 10 Video Solution by Power Solve (easy to digest!)
- 11 Video Solution (⚡Under 4 min⚡)
- 12 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 13 Video Solution
- 14 Video Solution by Math-X (First understand the problem!!!)
- 15 See Also
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
Solution 1 (Manipulation)
Let be the length of the shorter leg and be the longer leg. By the Pythagorean theorem, we can derive that . Using area we can also derive that . as given in the diagram, we can find that because . This means that and . Adding the equations gives and when is plugged in . Rationalizing the denominators gives us .
Solution 2 (Area)
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 side length 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
Solution 4
The side length of the bigger square is equal to , while the side length of the smaller square is . Let be the shorter leg and be the longer one. Clearly, , and . Using Vieta's to build a quadratic, we get Solving, we get and . Thus, we find .
~vadava_lx
Solution 5
Let be the angle opposite the smaller leg. We want to find .
The area of the triangle is which implies or . Therefore
Solution 6
Allow a, b to be the sides of a triangle. WLOG, suppose We want to find . Notice that the area of a triangle is , which results in . Thus, . However, the square of the hypotenuse of this triangle is , but also . We can write as , and then plug it in. We get , so . Applying the quadratic formula, , or . However, since and must both be solutions of the quadratic, since both equations were cyclic. Since , then , and . To find , we can simply find the square root of . This is , so the answer is . - Sepehr2010
Solution 6
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
Video Solution by Power Solve (easy to digest!)
https://www.youtube.com/watch?v=7KXT1pI-i64
Video Solution (⚡Under 4 min⚡)
~Education, the Study of Everything
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=IVgzVS86Ogo
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/N2lyYRMuZuk?si=99Ir3nUQxTkJDiVm
~Math-X
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.