Difference between revisions of "1986 AIME Problems/Problem 15"
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We get the quadratic <math>2r^2 -9r+2=0</math>, which solves to give <math>r=\frac{9 \pm \sqrt{65}}{4}</math>. It does not matter which one is picked, because the two roots multiply to 1, so switching from one root to another is like switching the lengths of AC and BC. | We get the quadratic <math>2r^2 -9r+2=0</math>, which solves to give <math>r=\frac{9 \pm \sqrt{65}}{4}</math>. It does not matter which one is picked, because the two roots multiply to 1, so switching from one root to another is like switching the lengths of AC and BC. | ||
− | We choose the positive root and we can plug <math>y=(\frac{9 + \sqrt{65}}{4})x</math> into of course the PYTHAGOREAN THEOREM <math>x^2 + y^2 =3600</math> and solve for <math>x^2=200(9-\sqrt{65})</math>. We want the area which is <math>\frac{xy}{2}=\frac{x^2(\frac{9 + \sqrt{65}}{4})}{2}=\frac{200(9-\sqrt{65})(\frac{9 + \sqrt{65}}{4})}{2}=\frac{200\frac{16}{4}}{2}=400</math>. | + | We choose the positive root and we can plug <math>y=(\frac{9 + \sqrt{65}}{4})x</math> into of course the PYTHAGOREAN THEOREM <math>x^2 + y^2 =3600</math> and solve for <math>x^2=200(9-\sqrt{65})</math>. We want the area which is <math>\frac{xy}{2}=\frac{x^2(\frac{9 + \sqrt{65}}{4})}{2}=\frac{200(9-\sqrt{65})(\frac{9 + \sqrt{65}}{4})}{2}=\frac{200\cdot\frac{16}{4}}{2}=400</math>. |
== See also == | == See also == |
Revision as of 23:01, 26 December 2019
Contents
[hide]Problem
Let triangle be a right triangle in the xy-plane with a right angle at . Given that the length of the hypotenuse is , and that the medians through and lie along the lines and respectively, find the area of triangle .
Solution 1
Translate so the medians are , and , then model the points and . is the centroid, and is the average of the vertices, so
so
and are perpendicular, so the product of their slopes is , giving
Combining and , we get
Using the determinant product for area of a triangle (this simplifies nicely, add columns 1 and 2, add rows 2 and 3), the area is , so we get the answer to be .
Solution 2
The only relevant part about the xy plane here is that the slopes of the medians determine an angle between them that we will use. This solution uses the tangent subtraction identity . Therefore, the tangent of the acute angle between the medians from A and B will be .
Let and . Let the midpoint of AC be M and the midpoint of BC be N. Let the centroid be G. By exterior angles, . However we know that since is the acute angle formed by the medians, . We can express the tangents of the other two angles in terms of and . while . For simplification, let .
By the tangent subtraction identity, . We get the quadratic , which solves to give . It does not matter which one is picked, because the two roots multiply to 1, so switching from one root to another is like switching the lengths of AC and BC.
We choose the positive root and we can plug into of course the PYTHAGOREAN THEOREM and solve for . We want the area which is .
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.