Difference between revisions of "1985 AHSME Problems/Problem 28"
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==Problem== | ==Problem== | ||
− | In <math> \triangle ABC </math>, we have <math> \angle C=3\angle A, a=27 | + | In <math>\triangle ABC</math>, we have <math>\angle C = 3\angle A</math>, <math>a = 27</math> and <math>c = 48</math>. What is <math>b</math>? |
<asy> | <asy> | ||
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<math> \mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined} </math> | <math> \mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined} </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | Let <math>\angle A = x^{\circ}</math>, so <math>\angle C = 3x^{\circ}</math>, and thus <math>\angle B = \left(180-4x\right)^{\circ}</math>. Now let <math>D</math> be a point on side <math>AB</math> such that <math>\angle ACD = x^{\circ}</math>, so <math>\angle BCD = 3x^{\circ}-x^{\circ} = 2x^{\circ}</math>, which gives <cmath>\angle CDB = 180^{\circ}-2x^{\circ}-\left(180-4x\right)^{\circ} = 2x^{\circ},</cmath> meaning that <math>\triangle CDB</math> and <math>\triangle CDA</math> are both isosceles, with <math>BC = BD</math> and <math>AD = CD</math>. In particular, <math>BD = BC = 27</math> and <math>CD = AD = AB-BD = 48-27 = 21</math>. Hence by [[Stewart's theorem]] on triangle <math>ABC</math>, <cmath>\begin{align*}&BD \cdot AB \cdot AD + CD^2 \cdot AB = AC^2 \cdot BD + BC^2 \cdot AD \\ &\iff 27 \cdot 48 \cdot 21 + 21^2 \cdot 48 = AC^2 \cdot 27 + 27^2 \cdot 21 \\ &\iff AC^2 = \frac{27(21)(48-27) + 21^2 \cdot 48}{27} \\ &\iff AC^2 = \frac{21^2(27+48)}{27} \\ &\iff AC^2 = \frac{21^2 \cdot 25}{9} \\ &\iff AC = \frac{21 \cdot 5}{3} \qquad \text{(as } AC > 0\text{)} \\ &\iff AC = \boxed{\text{(B)} \ 35}.\end{align*}</cmath> | |
− | We | + | ==Solution 2== |
+ | We apply the law of sines in the form <cmath>\frac{\sin(A)}{a} = \frac{\sin(C)}{c},</cmath> yielding <cmath>\frac{\sin(A)}{27} = \frac{\sin(3A)}{48} \iff 9\sin(3A) = 16\sin(A).</cmath> | ||
+ | Now, the angle sum and double angle identities give <cmath>\begin{align*}\sin(3A) &= \sin(2A+A) \\ &= \sin(2A)\cos(A)+\cos(2A)\sin(A) \\ &= \left(2\sin(A)\cos(A)\right)\cos(A)+\left(\cos^2(A)-\sin^2(A)\right)\sin(A) \\ &= 2\sin(A)\cos^2(A)+\sin(A)\cos^2(A)-\sin^3(A) \\ &= 3\sin(A)\left(1-\sin^2(A)\right)-\sin^3(A) \\ &\text{(using the further identity } \cos^2(\theta)+\sin^2(\theta) = 1\text{)} \\ &= 3\sin(A)-4\sin^3(A).\end{align*}</cmath> | ||
− | < | + | Thus our equation becomes <cmath>\begin{align*}9\left(3\sin(A)-4\sin^3(A)\right) = 16\sin(A) &\iff 27\sin(A)-36\sin^3(A) = 16\sin(A) \\ &\iff 36\sin^3(A) = 11\sin(A) \\ &\iff \sin(A) = 0 \text{ or } \pm\frac{\sqrt{11}}{6}.\end{align*}</cmath> |
− | + | Notice, however, that we must have <math>0^{\circ} < A < 45^{\circ}</math>, the latter because otherwise <math>A+3A \geq 180^{\circ}</math>, which would contradict the fact that <math>A</math> and <math>3A</math> are angles in a (non-degenerate) triangle. This means <math>\sin(A) > 0</math>, so the only valid solution is <cmath>\sin(A) = \frac{\sqrt{11}}{6},</cmath> and the fact that <math>A</math> is acute also means <math>\cos(A) > 0</math>, so we deduce <cmath>\begin{align*}\cos(A) &= \sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2} \\ &= \sqrt{1-\frac{11}{36}} \\ &=\frac{5}{6}.\end{align*}</cmath> | |
− | + | Accordingly, using the double angle identities again, <cmath>\begin{align*}\sin(4A) &= \sin(2 \cdot 2A) \\ &= 2\sin(2A)\cos(2A) \\ &= 2\left(2\sin(A)\cos(A)\right)\left(\cos^2(A)-\sin^2(A)\right) \\ & =2\left(2 \cdot \frac{\sqrt{11}}{6} \cdot\frac{5}{6}\right)\left(\left(\frac{5}{6}\right)^2-\left(\frac{\sqrt{11}}{6}\right)^2\right) \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{25-11}{36} \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{7}{18} \\ &= \frac{35\sqrt{11}}{162}.\end{align*}</cmath> | |
− | + | Finally, the law of sines now gives <cmath>\begin{align*}\frac{\sin(A)}{27} &= \frac{\sin(B)}{b} \\ &= \frac{\sin(180^{\circ}-3A-A)}{b} \\ &= \frac{\sin(4A)}{b} \qquad \text{(using the identity } \sin\left(180^{\circ}-\theta\right) = \sin(\theta)\text{)},\end{align*}</cmath> so, substituting the above results, <cmath>\frac{\left(\frac{\sqrt{11}}{6}\right)}{27} = \frac{\left(\frac{35\sqrt{11}}{162}\right)}{b} \iff b = \frac{6 \cdot 27 \cdot 35}{162} = \boxed{\text{(B)} \ 35}.</cmath> | |
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==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=27|num-a=29}} | {{AHSME box|year=1985|num-b=27|num-a=29}} | ||
+ | {{MAA Notice}} |
Latest revision as of 02:54, 20 March 2024
Contents
Problem
In , we have
,
and
. What is
?
Solution 1
Let , so
, and thus
. Now let
be a point on side
such that
, so
, which gives
meaning that
and
are both isosceles, with
and
. In particular,
and
. Hence by Stewart's theorem on triangle
,
Solution 2
We apply the law of sines in the form yielding
Now, the angle sum and double angle identities give
Thus our equation becomes
Notice, however, that we must have
, the latter because otherwise
, which would contradict the fact that
and
are angles in a (non-degenerate) triangle. This means
, so the only valid solution is
and the fact that
is acute also means
, so we deduce
Accordingly, using the double angle identities again,
Finally, the law of sines now gives
so, substituting the above results,
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.