Difference between revisions of "2021 AMC 12B Problems/Problem 24"
Skyscraper (talk | contribs) (→Solution 5) |
Skyscraper (talk | contribs) (→Solution 5) |
||
Line 36: | Line 36: | ||
Substituting in our values, | Substituting in our values, | ||
<cmath> \frac{\frac{15}{2(x+3)}}{\sqrt{9+(\frac{15}{2(x+3)})^2}} = \frac{\sqrt{(x+3)^2 - 4^2}}{x+3}</cmath> | <cmath> \frac{\frac{15}{2(x+3)}}{\sqrt{9+(\frac{15}{2(x+3)})^2}} = \frac{\sqrt{(x+3)^2 - 4^2}}{x+3}</cmath> | ||
− | To rid unnecessary computation, we let <math>x+3 = a.</math> The equation simplifies, after cross multiplying, to | + | To rid unnecessary computation, we let <math>(x+3)^2 = a.</math> The equation simplifies, after cross multiplying, to |
<cmath> \sqrt{9+\frac{15^2}{4a}} \sqrt{a-16 } = \frac{15}{2} </cmath> | <cmath> \sqrt{9+\frac{15^2}{4a}} \sqrt{a-16 } = \frac{15}{2} </cmath> | ||
<cmath> 36a^2 - 576a - 15^2\cdot 16 = 0</cmath> | <cmath> 36a^2 - 576a - 15^2\cdot 16 = 0</cmath> |
Latest revision as of 16:15, 14 October 2021
Contents
Problem
Let be a parallelogram with area . Points and are the projections of and respectively, onto the line and points and are the projections of and respectively, onto the line See the figure, which also shows the relative locations of these points.
Suppose and and let denote the length of the longer diagonal of Then can be written in the form where and are positive integers and is not divisible by the square of any prime. What is
Solution 1
Let denote the intersection point of the diagonals and . Remark that by symmetry is the midpoint of both and , so and . Now note that since , quadrilateral is cyclic, and so which implies .
Thus let be such that and . Then Pythagorean Theorem on yields , and soSolving this for yields , and soThe requested answer is .
Solution 2 (Trig)
Let denote the intersection point of the diagonals and and let . Then, by the given conditions, . So, Combining the above 3 equations, we get Since we want to find we let Then Solving this, we get so .
Solution 3 (Similar Triangles and Algebra)
Let be the intersection of diagonals and . By symmetry , and , so now we have reduced all of the conditions one quadrant. Let . , by similar triangles and using the area condition we get . Note that it suffices to find because we can double and square it to get . Solving for in the above equation, and then using .
Solution 4 (Similar Triangles)
Again, Let be the intersection of diagonals and . Note that triangles and are similar because they are right triangles and share . First, call the length of . By the definition of an area of a parallelogram, , so . Using similar triangles on and , . Therefore, finding , . Now, applying the Pythagorean theorem once, we find + = . Solving this equation for , we find .
Solution 5
Let We know that the area of the parallelogram is so it follows that and the height of each triangle, which are also the lengths of and is Suppose that Because and we have The length of by the Pythagorean Theorem is and the length of by the Pythagorean Theorem on is Note that Substituting in our values, To rid unnecessary computation, we let The equation simplifies, after cross multiplying, to By the quadratic formula, so we discard the negative solution. The value of is and the desired answer is ~skyscraper
Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point)
~ pi_is_3.14
Video Solution (Simple: Using trigonometry and Equations)
https://youtu.be/ZB-VN02H6mU ~hippopotamus1
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.