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| | and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>. | | and the answer is <math>a+b+c = 216 + 72 + 3 = \boxed{291}</math>. |
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| − | ==Solution 2==
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| − | Since <math>\angle CAB</math> has a measure of <math>60^{\circ}</math>, and thus has sines and cosines that are easy to compute, we attempt to find <math>AC</math> and <math>AB</math>, and use the formula that
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| − | <math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math>
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| − | By angle chasing, we find that <math>ACT</math> is a triangle with <math>\angle A = 30^{\circ}, \angle C = 75^{\circ}</math> and <math>\angle T = 75^{\circ}</math>. Thus <math>AC = AT = 24</math>.
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| − | Switching to the lower triangle <math>ATB</math>, <math>\angle A = 30^{\circ}, \angle T = 105^{\circ}</math>, and <math>\angle B = 45^{\circ}</math>, with <math>AT = 24</math>.
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| − | Using the [[Law of Sines]] on <math>ATB</math>:
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| − | <math>\frac{AT}{\sin 45^{\circ}} = \frac{AB}{\sin 105^{\circ}}</math>
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| − | <math>24 \cdot \sqrt{2} \cdot \sin 105^{\circ} = AB</math>
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| − | <math>24 \cdot \sqrt{2} \cdot \sin (60^{\circ}+ 45^{\circ}) = AB</math>
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| − | <math>24 \cdot \sqrt{2} \cdot (\sin 60^{\circ} \cos 45^{\circ} + \sin 45^{\circ} \cos 60^{\circ}) = AB</math>
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| − | <math>24 \cdot \sqrt{2} \cdot (\frac {\sqrt{3}}{2} + \frac{1}{2}) \cdot \frac{\sqrt{2}}{2} = AB</math>
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| − | <math>AB = 12 \cdot (\sqrt{3} + 1)</math>
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| − | We now plug in <math>AC</math>, <math>AB</math> and <math>\sin \angle A</math> into the formula for the area:
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| − | <math>[ABC] = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \angle A</math>
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| − | <math>[ABC] = \frac{1}{2} \cdot 24 \cdot 12 \cdot (\sqrt{3} + 1) \cdot \frac{\sqrt{3}}{2}</math>
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| − | <math>[ABC] = 72\sqrt{3} \cdot (\sqrt{3} + 1)</math>
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| − | <math>[ABC] = 216 + 72\sqrt{3}</math>
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| − | Thus the answer is <math>216 + 72 + 3 = \boxed{291}</math>
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| − | Note: We could also get the lengths (and area) of the triangle by drawing a perpendicular from <math>T</math> to <math>AB</math>, forming a <math>30-60-90</math> and a <math>45-45-90</math> triangle.
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| | == See also == | | == See also == |
Problem
In triangle 
, angles 
and 
measure 
degrees and 
degrees, respectively. The bisector of angle intersects 
at 
, and 
. The area of triangle can be written in the form 
, where 
, 
, and 
are positive integers, and is not divisible by the square of any prime. Find 
.
Solution 1
![[asy] size(180); pointpen = black; pathpen = black+linewidth(0.7); pair A=(0,0),B=(12+12*3^.5,0),C=(12,12*3^.5),D=foot(C,A,B),T=IP(CR(A,24),B--C); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(D(MP("T",T,NE))--A); D(MP("D",D)--C,linetype("6 6") + linewidth(0.7)); MP("24",(A+3*T)/4,SE); D(anglemark(C,B,A,65)); D(anglemark(B,A,C,65)); D(rightanglemark(C,D,B,50)); MP("30^{\circ}",A,(4,1)); MP("45^{\circ}",B,(-3,1)); [/asy]](//latex.artofproblemsolving.com/f/d/8/fd80ef0be57cda8d5c014431604ac07fc2c9a044.png)
Let 
be the foot of the altitude from 
to 
. By simple angle-chasing, we find that 
, and thus 
. Now 
is a 
right triangle and 
is a 
right triangle, so 
. The area of
![\[ABC = \frac{1}{2}bh = \frac{CD \cdot (AD + BD)}{2} = \frac{12\sqrt{3} \cdot \left(12\sqrt{3} + 12\right)}{2} = 216 + 72\sqrt{3},\]](//latex.artofproblemsolving.com/0/7/4/07457be2b941b036eb23b9ba145d240d422a3b25.png)
and the answer is 
.
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
, angles 
and 
measure 
degrees and 
degrees, respectively. The bisector of angle 
at 
, and 
. The area of triangle 
, where 
, 
, and 
are positive integers, and 
.
![[asy] size(180); pointpen = black; pathpen = black+linewidth(0.7); pair A=(0,0),B=(12+12*3^.5,0),C=(12,12*3^.5),D=foot(C,A,B),T=IP(CR(A,24),B--C); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(D(MP("T",T,NE))--A); D(MP("D",D)--C,linetype("6 6") + linewidth(0.7)); MP("24",(A+3*T)/4,SE); D(anglemark(C,B,A,65)); D(anglemark(B,A,C,65)); D(rightanglemark(C,D,B,50)); MP("30^{\circ}",A,(4,1)); MP("45^{\circ}",B,(-3,1)); [/asy]](http://latex.artofproblemsolving.com/f/d/8/fd80ef0be57cda8d5c014431604ac07fc2c9a044.png)
be the foot of the altitude from 
to 
. By simple angle-chasing, we find that 
, and thus 
. Now 
is a 
right triangle and 
is a 
right triangle, so 
. The area of
![\[ABC = \frac{1}{2}bh = \frac{CD \cdot (AD + BD)}{2} = \frac{12\sqrt{3} \cdot \left(12\sqrt{3} + 12\right)}{2} = 216 + 72\sqrt{3},\]](http://latex.artofproblemsolving.com/0/7/4/07457be2b941b036eb23b9ba145d240d422a3b25.png)
.
