2022 AMC 10B Problems/Problem 16
- The following problem is from both the 2022 AMC 10B #16 and 2022 AMC 12B #13, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Clever)
- 4 Solution 3 (Coordinate Geometry)
- 5 Solution 4
- 6 Solution 5 (Fastest Similar Triangles)
- 7 Video Solution by mop 2024
- 8 Video Solution
- 9 Video Solution by SpreadTheMathLove
- 10 Video Solution 3 by OmegaLearn
- 11 Video Solution(1-16)
- 12 Video Solution by Interstigation
- 13 See Also
Problem
The diagram below shows a rectangle with side lengths and and a square with side length . Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?
Solution 1
Let us label the points on the diagram.
By doing some angle chasing using the fact that and are right angles, we find that . Similarly, . Therefore, .
As we are given a rectangle and a square, and . Therefore, is a -- right triangle and .
is also . So, using the similar triangles, and .
. Using the similar triangles again, is of the corresponding . So,
Finally, we have
~Connor132435
Solution 2 (Clever)
(Refer to the diagram above) Proceed the same way as Solution 1 until you get all of the side lengths. Then, it is clear that due to the answer choices, we only need to find the fractional part of the shaded area. The area of the whole rectangle is integral, as is the area of , , and the rectangle to the far left of the diagram. The area of is and thus the fractional part of the answer is . The only answer choice that has in it is
~mathboy100
Solution 3 (Coordinate Geometry)
Same diagram as Solution 1, but added point , which is . I also renamed all the points to form coordinates using as the origin.
In order to find the area, point 's coordinates must be found. Notice how and intercept at point . This means that we need to find the equations for and and make a system of linear equations.
Using the slope formula , we get the slope for , which means
Then, by using point slope form. . We can say that the equation for is or in this case, .
And it is easy to figure out that the equation for is .
The best way to solve the system of linear equations is to substitute the for the in equation . , so and This would mean .
Since we have our coordinate, we can continue with Solution 3, with the area of the trapezoid , where (using distance formula for to ), , and .
By substitution, we get .
~ghfhgvghj10 (+ minor edits ~TaeKim)
Solution 4
Notice the small triangle in the upper right corner is a triangle. Then that triangle is similar to the big triangle by AA similarity. From that, do similar triangles and you find that the longer leg of the small triangle is 1. Then you find that the triangle below is , so the side length of the rectangle (without the outer rectangle) is 7. afterwards you just add the half of the square + the remaining triangle which can be found by multiplying base and height (in which we already know)
~mathboy282
Solution 5 (Fastest Similar Triangles)
For reference, use the points labelled in the diagram of Solution 3. The square means , so we get a -- triangle .
.
Therefore is proportional to a -- triangle, with corresponding to and corresponding to . By similar triangles, we find
.
Finally,
~lolsmybagelz
Video Solution by mop 2024
https://youtu.be/ezGvZgBLe8k&t=347s
~r00tsOfUnity
Video Solution
~Education, the Study of Everything
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=NAu4AKlK-O0&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=2
~Ismail.maths
Video Solution 3 by OmegaLearn
~ pi_is_3.14
Video Solution(1-16)
~~Hayabusa1
Video Solution by Interstigation
~Interstigation
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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.