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Difference between revisions of "1985 AHSME Problems/Problem 28"

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==Problem==
 
==Problem==
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In <math> \triangle ABC </math>, we have <math> \angle C=3\angle A, a=27 </math> and <math> c=48 </math>. What is <math> b </math>?
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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>?
  
 
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<asy>
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==Solution 1==
 
==Solution 1==
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From the Law of Sines, we have <math> \frac{\sin(A)}{a}=\frac{\sin(C)}{c} </math>, or <math> \frac{\sin(A)}{27}=\frac{\sin(3A)}{48}\implies 9\sin(3A)=16\sin(A) </math>.
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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>
  
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We now need to find an identity relating <math> \sin(3A) </math> and <math> \sin(A) </math>. We have
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==Solution 2==
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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>
  
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<math> \sin(3A)=\sin(2A+A)=\sin(2A)\cos(A)+\cos(2A)\sin(A) </math>
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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>
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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>
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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>
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<math> =2\sin(A)\cos^2(A)+\sin(A)\cos^2(A)-\sin^3(A)=3\sin(A)\cos^2(A)-\sin^3(A) </math>
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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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<math> =3\sin(A)(1-\sin^2(A))-\sin^3(A)=3\sin(A)-4\sin^3(A) </math>.
 
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Thus we have <math> 9(3\sin(A)-4\sin^3(A))=27\sin(A)-36\sin^3(A)=16\sin(A) </math>
 
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<math> \implies 36\sin^3(A)=11\sin(A) </math>.
 
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Therefore, <math> \sin(A)=0, \frac{\sqrt{11}}{6}, </math> or <math> -\frac{\sqrt{11}}{6} </math>. Notice that we must have <math> 0^\circ<A<45^\circ </math> because otherwise <math> A+3A>180^\circ </math>. We can therefore disregard <math> \sin(A)=0 </math> because then <math> A=0 </math> and also we can disregard <math> \sin(A)=-\frac{\sqrt{11}}{6} </math> because then <math> A </math> would be in the third or fourth quadrants, much greater than the desired range.
 
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Therefore, <math> \sin(A)=\frac{\sqrt{11}}{6} </math>, and <math> \cos(A)=\sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2}=\frac{5}{6} </math>. Going back to the Law of Sines, we have <math> \frac{\sin(A)}{27}=\frac{\sin(B)}{b}=\frac{\sin(\pi-3A-A)}{b}=\frac{\sin(4A)}{b} </math>.
 
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We now need to find <math> \sin(4A) </math>.
 
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<math> \sin(4A)=\sin(2\cdot2A)=2\sin(2A)\cos(2A) </math>
 
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<math> =2(2\sin(A)\cos(A))(\cos^2(A)-\sin^2(A)) </math>
 
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<math> =2\cdot2\cdot\frac{\sqrt{11}}{6}\cdot\frac{5}{6}\left(\left(\frac{5}{6}\right)^2-\left(\frac{\sqrt{11}}{6}\right)^2\right)=\frac{35\sqrt{11}}{162} </math>.
 
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Therefore, <math> \frac{\frac{\sqrt{11}}{6}}{27}=\frac{\frac{35\sqrt{11}}{162}}{b}\implies b=\frac{6\cdot27\cdot35}{162}=35, \boxed{\text{B}} </math>.
 
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== Solution 2 ==
 
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Let angle <math>A</math> be equal to <math>x</math> degrees. Then angle <math>C</math> is equal to <math>3x</math> degrees, and angle <math>B</math> is equal to <math>180-4x</math> degrees. Let <math>D</math> be a point on side <math>AB</math> such that <math>\angle ACD</math> is equal to <math>x</math> degrees. Because <math>2x+180-4x+\angle CDB=180</math>, angle <math>CDB</math> is equal to <math>2x</math> degrees. We can now see that triangles <math>CDB</math> and <math>CDA</math> are both isosceles, with <math>CB=DB</math> and <math>AD=AC</math>. From isosceles triangle <math>CDB</math>, we now know that <math>BD = 27</math>, and since <math>AB = c = 48</math>, we know that <math>AD = 21</math>. From isosceles triangle <math>CDA</math>, we now know that <math>CD = 21</math>. Applying [[Stewart's Theorem]] on triangle <math>ABC</math> gives us <math>AC = 35</math>, which is <math>\boxed{\text{B}}</math>.
 
  
 
==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 02:54, 20 March 2024

Problem

In $\triangle ABC$, we have $\angle C = 3\angle A$, $a = 27$ and $c = 48$. What is $b$?

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C--A)*dir(-90)); label("$c$", A--B, dir(A--B)*dir(-90));[/asy]

$\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined}$

Solution 1

Let $\angle A = x^{\circ}$, so $\angle C = 3x^{\circ}$, and thus $\angle B = \left(180-4x\right)^{\circ}$. Now let $D$ be a point on side $AB$ such that $\angle ACD = x^{\circ}$, so $\angle BCD = 3x^{\circ}-x^{\circ} = 2x^{\circ}$, which gives \[\angle CDB = 180^{\circ}-2x^{\circ}-\left(180-4x\right)^{\circ} = 2x^{\circ},\] meaning that $\triangle CDB$ and $\triangle CDA$ are both isosceles, with $BC = BD$ and $AD = CD$. In particular, $BD = BC = 27$ and $CD = AD = AB-BD = 48-27 = 21$. Hence by Stewart's theorem on triangle $ABC$, \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*}

Solution 2

We apply the law of sines in the form \[\frac{\sin(A)}{a} = \frac{\sin(C)}{c},\] yielding \[\frac{\sin(A)}{27} = \frac{\sin(3A)}{48} \iff 9\sin(3A) = 16\sin(A).\]

Now, the angle sum and double angle identities give \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*}

Thus our equation becomes \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*} Notice, however, that we must have $0^{\circ} < A < 45^{\circ}$, the latter because otherwise $A+3A \geq 180^{\circ}$, which would contradict the fact that $A$ and $3A$ are angles in a (non-degenerate) triangle. This means $\sin(A) > 0$, so the only valid solution is \[\sin(A) = \frac{\sqrt{11}}{6},\] and the fact that $A$ is acute also means $\cos(A) > 0$, so we deduce \begin{align*}\cos(A) &= \sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2} \\ &= \sqrt{1-\frac{11}{36}} \\ &=\frac{5}{6}.\end{align*} Accordingly, using the double angle identities again, \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*} Finally, the law of sines now gives \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*} so, substituting the above results, \[\frac{\left(\frac{\sqrt{11}}{6}\right)}{27} = \frac{\left(\frac{35\sqrt{11}}{162}\right)}{b} \iff b = \frac{6 \cdot 27 \cdot 35}{162} = \boxed{\text{(B)} \ 35}.\]

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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