2021 Fall AMC 10B Problems/Problem 15
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Similarity, Pythagorean Theorem, and Systems of Equations)
- 4 Solution 3
- 5 Solution 4 (Point-line distance formula)
- 6 Solution 5
- 7 Solution 6 (Answer choices and areas)
- 8 Solution 7 (Power of a Point)
- 9 Video Solution by Interstigation
- 10 Video Solution
- 11 Video Solution by WhyMath
- 12 Video Solution by TheBeautyofMath
- 13 Video Solution by OmegaLearn
- 14 See Also
Problem
In square , points and lie on and , respectively. Segments and intersect at right angles at , with and . What is the area of the square?
Solution 1
Note that by ASA. ( and ) Then, it follows that Thus, Define to be the length of side then Because is the altitude of the triangle, we can use the property that Substituting the given lengths, we have Solving, gives and We eliminate the possibility of because Thus, the side length of the square, by Pythagorean Theorem, is Thus, the area of the square is so the answer is
Note that there is another way to prove that is impossible. If then the side length would be and the area would be but that isn't in the answer choices. Thus, must be
~NH14 ~sl_hc
Extra Note: Another way to prove is impossible. The side length of the square, , is equal to . Because , . Because and but , we have proof by contradiction. And so .
~ Wiselion (Extra Note)
Solution 2 (Similarity, Pythagorean Theorem, and Systems of Equations)
As above, note that , which means that . In addition, note that is the altitude of a right triangle to its hypotenuse, so . Let the side length of the square be ; using similarity side ratios of to , we get Note that by the Pythagorean theorem, so we can use the expansion to produce two equations and two variables;
Since is negative, it doesn't make sense in the context of this problem, so we go with
We want , so we want to find . Adding the first equation to the second, we get
Then =
~KingRavi
~stjwyl (Edits)
-yingkai_0_ (Minor Edits)
Solution 3
We have that Thus, . Now, let the side length of the square be Then, by the Pythagorean theorem, Plugging all of this information in, we get Simplifying gives Squaring both sides gives We now set and get the equation From here, notice we want to solve for , as it is precisely or the area of the square. So we use the Quadratic formula, and though it may seem bashy, we hope for a nice cancellation of terms. It seems scary, but factoring from the square root gives us giving us the solutions We instantly see that is way too small to be an area of this square ( isn't even an answer choice, so you can skip this step if out of time) because then the side length would be and then, even the largest line you can draw inside the square (the diagonal) is which is less than (line ) And thus, must be , and our answer is
~wamofan
Solution 4 (Point-line distance formula)
Denote . Now tilt your head to the right and view and as the origin, -axis and -axis, respectively. In particular, we have points . Note that side length of the square is . Also equation of line is Because the distance from to line is also the side length , we can apply the point-line distance formula to get which reduces to . Since is positive, the last equations factors as . Now judging from the figure, we learn that . So . Therefore, the area of the square is .
~VensL. can someone explain how line bc doesnt have an undefined slope?? how are there 2 different x coordinates if its a vertical line... ~THATONEKID_-TOKYT- The line is not vertical. Looking closely, you can actually see that CQ and PB lie on the Y and X axis respectively.
Solution 5
Denote . Because , .
Hence, , .
Because is a square, . Hence, .
Therefore,
Thus, .
: .
Thus, .
Hence, .
Therefore, .
: .
Thus, .
Hence, .
However, we observe . Therefore, in this case, point is not on the segment .
Therefore, this case is infeasible.
Putting all cases together, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 6 (Answer choices and areas)
Note that if we connect points and , we get a triangle with height and length . This triangle has an area of the square. We can now use answer choices to our advantage!
Answer choice A: If was , would be . The triangle would therefore have an area of which is not half of the area of the square. Therefore, A is wrong.
Answer choice B: If was , would be . This is obviously wrong.
Answer choice C: If was , we would have that is . The area of the triangle would be , which is not half the area of the square. Therefore, C is wrong.
Answer choice D: If was , that would mean that is . The area of the triangle would therefore be which IS half the area of the square. Therefore, our answer is .
~Arcticturn
Solution 7 (Power of a Point)
Note that is a cyclic quadrilateral (opposite angles add to ). Call the circumcircle of quadrilateral . Then the power of to is . Let be the length of and the side length of the square, then we have , and we also have , solving the two equation will give us .
~student99
~minor edits by Yiyj1
Video Solution by Interstigation
https://www.youtube.com/watch?v=sKC0Yt6sPi0
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/R7TwXgAGYuw?t=1367 (note in the comments an easier solution too from a viewer)
~IceMatrix
Video Solution by OmegaLearn
https://youtu.be/hDsoyvFWYxc?t=822
~ pi_is_3.14
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.