2017 AMC 12B Problems/Problem 24
Contents
Problem
Quadrilateral has right angles at and , , and . There is a point in the interior of such that and the area of is times the area of . What is ?
Solution 1
Let , , and . Note that . By the Pythagorean Theorem, . Since , the ratios of side lengths must be equal. Since , and . Let F be a point on such that is an altitude of triangle . Note that . Therefore, and . Since and form altitudes of triangles and , respectively, the areas of these triangles can be calculated. Additionally, the area of triangle can be calculated, as it is a right triangle. Solving for each of these yields:
\begin{align*} [BEC] &=[CED]=[BEA]=\frac{x^3}{2(x^2+1)} \\ [ABCD] &=[AED]+[DEC]+[CEB]+[BEA] \\ \frac{(BC)(AB+CD)}{2} &= 17*[CEB]+ [CEB] + [CEB] + [CEB] \\ \frac{x^3+x}{2} &= \frac{20x^3}{2(x^2+1)} \\ \frac{x}{x^2+1} &= \frac{20x^3}{x^2+1} \\ (x^2+1)^2 &=20x^2 \\ x^4-18x^2+1 &=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5 \\ \end{align*}
Therefore, the answer is
Solution 2
Draw line through , with on and on , . WLOG let , , . By weighted average .
Meanwhile, . This follows from comparing the ratios of triangle DEG to CFE and triangle AEG to FEB, both pairs in which the two triangles share a height perpendicular to FG, and have base ratio .
. We obtain , namely .
The rest is the same as Solution 1.
Solution 3
Let , ,
Note that cannot be the intersection of and , as that would mean
Let ,
Solution 4
Let . Then from the similar triangles condition, we compute and . Hence, the -coordinate of is just . Since lies on the unit circle, we can compute the coordinate as . By Shoelace, we want Factoring out denominators and expanding by minors, this is equivalent to This factors as , so and so the answer is .
Solution 5
Let where . Because . Notice that the diagonals are perpendicular with slopes of and . Let the intersection of and be , then . However, because is a trapezoid, and share the same area, therefore is the reflection of over the perpendicular bisector of , which is . We use the linear equations of the diagonals, , to find the coordinates of . The y-coordinate of is simply The area of is . We apply shoelace theorem to solve for the area of . The coordinates of the triangle are , so the area is
Finally, we use the property that the ratio of areas equals
~Zeric
Solution 6
This solution involves proving .
Let be the intersection of and . Label points and the same way as .
. Additionally, , so by SAS. Therefore, .
Next, because . Also, , so . Therefore, by AA. Since , .
Given , we deduce that the ratio of corresponding side lengths of to must be . Now, we set , , and . Using the Pythagorean Theorem, . Thus, . Solving gives .
Finally, .
~Zhixing
Video Solution by MOP 2024
~r00tsOfUnity
Notes
1) is the most relevant answer choice because it shares numbers with the givens of the problem.
2) It's a very good guess to replace finding the area of triangle AED with the area of the triangle DAF, where F is the projection of D onto AB(then find the closest answer choice).
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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