Difference between revisions of "2002 AMC 12A Problems/Problem 23"
Isabelchen (talk | contribs) m (→Solution 4 (Trigonometry)) |
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We plug this to find that the sides of <math>\triangle ABD</math> are <math>12,7,9</math>. By Heron's formula, the area is <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. ~skyscraper | We plug this to find that the sides of <math>\triangle ABD</math> are <math>12,7,9</math>. By Heron's formula, the area is <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. ~skyscraper | ||
− | ==Solution 4 (Trigonometry) == | + | ==Solution 4== |
+ | |||
+ | <asy> | ||
+ | size(12cm, 12cm); | ||
+ | pair A, B, C, D, M, N; | ||
+ | A = (0,0); | ||
+ | B = (88/9, 28*sqrt(5)/9); | ||
+ | C = (16,0); | ||
+ | D = 9/16*C; | ||
+ | M = (B + C)/2; | ||
+ | N = (6,4.27); | ||
+ | |||
+ | draw(A--B--C--cycle); | ||
+ | draw(B--D--M); | ||
+ | draw(D--N--B); | ||
+ | label("$A$", A, SW); | ||
+ | label("$B$", B, N); | ||
+ | label("$C$", C, SE); | ||
+ | label("$D$", D, S); | ||
+ | label("$M$", M, NE); | ||
+ | label("$N$", N, NW); | ||
+ | |||
+ | draw(rightanglemark(B, M, D), linewidth(.5)); | ||
+ | draw(rightanglemark(A, N, D), linewidth(.5)); | ||
+ | </asy> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 5 (Trigonometry) == | ||
Let <math>\angle ACB = \theta</math>, <math>\angle DBC = \theta</math>, <math>\angle ABD = \theta</math>, <math>\angle ADB = 2 \theta</math>, <math>\angle BAC = 180^\circ - 3 \theta</math> | Let <math>\angle ACB = \theta</math>, <math>\angle DBC = \theta</math>, <math>\angle ABD = \theta</math>, <math>\angle ADB = 2 \theta</math>, <math>\angle BAC = 180^\circ - 3 \theta</math> |
Revision as of 11:00, 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
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.