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> | ||
− | < | + | 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> | |
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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}} | {{MAA Notice}} |
Latest revision as of 01:54, 20 March 2024
Contents
[hide]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 |
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