Difference between revisions of "2002 AMC 12A Problems/Problem 23"
Isabelchen (talk | contribs) (→Solution 4) |
Isabelchen (talk | contribs) m (→Solution 4) |
||
Line 76: | Line 76: | ||
</asy> | </asy> | ||
− | Draw <math>DN</math> such that <math>DN \bot AB</math>, <math>\triangle BND \cong \triangle BMD</math>. | + | Draw <math>DN</math> such that <math>DN \bot AB</math>, <math>\triangle BND \cong \triangle BMD</math>. |
<math>\angle ACB = \angle DBC = \angle ABD</math>, <math>\triangle ABD \sim \triangle ACB</math> by <math>AA</math> | <math>\angle ACB = \angle DBC = \angle ABD</math>, <math>\triangle ABD \sim \triangle ACB</math> by <math>AA</math> |
Revision as of 12:08, 26 August 2022
Contents
[hide]Problem
In triangle , side and the perpendicular bisector of meet in point , and bisects . If and , what is the area of triangle ?
Solution 1
Looking at the triangle , we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let , so that from given and the previous deducted. Then because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means and are similar, so .
Then by using Heron's Formula on (with sides ), we have .
Solution 2
Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, and . Also, by the angle bisector theorem, . Thus, let and . In addition, .
Thus, . Additionally, using the Law of Cosines and the fact that ,
Substituting and simplifying, we get
Thus, . We now know all sides of . Using Heron's Formula on ,
Solution 3
Note that because the perpendicular bisector and angle bisector meet at side and as triangle is isosceles, so . By the angle bisector theorem, we can express and as and respectively. We try to find through Stewart's Theorem. So
We plug this to find that the sides of are . By Heron's formula, the area is . ~skyscraper
Solution 4
Draw such that , .
, by
, ,
By the Angle Bisector Theorem,
, ,
Solution 5 (Trigonometry)
Let , , , ,
By the Law of Sines we have
By the Triple-angle Identities,
, ,
,
By the Double Angle Identity
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.