Difference between revisions of "2006 iTest Problems/Problem 32"

(Solution to Problem 32 -- Triangle Angle Trisection)
 
m (Made fractions of final step bigger)
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&= \frac{84\sqrt{10}}{5}
 
&= \frac{84\sqrt{10}}{5}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
Thus, <math>\tfrac{AB}{AC} = \tfrac{84\sqrt{10}}{5} \cdot \tfrac{1}{180} = \tfrac{7\sqrt{10}}{75}</math>, so <math>p+q+r = \boxed{92}</math>.  
+
Thus, <math>\frac{AB}{AC} = \frac{84\sqrt{10}}{5} \cdot \frac{1}{180} = \frac{7\sqrt{10}}{75}</math>, so <math>p+q+r = \boxed{92}</math>.  
  
 
==See Also==
 
==See Also==

Revision as of 14:50, 15 December 2018

Problem

Triangle $ABC$ is scalene. Points $P$ and $Q$ are on segment $BC$ with $P$ between $B$ and $Q$ such that $BP=21$, $PQ=35$, and $QC=100$. If $AP$ and $AQ$ trisect $\angle A$, then $\tfrac{AB}{AC}$ can be written uniquely as $\tfrac{p\sqrt q}r$, where $p$ and $r$ are relatively prime positive integers and $q$ is a positive integer not divisible by the square of any prime. Determine $p+q+r$.

Solution

Let $a = AB$ and $b = AC$. Since $\angle BAP = \angle PAQ = \angle QAC$, by the Angle Bisector Theorem, we have $AP = \tfrac{7}{20}b$ and $AQ = \tfrac{5}{3}a$.


By using the Law of Cosines on $\triangle BAP$ and $\triangle PAQ$, we have \begin{align*} \frac{a^2 + \frac{49}{400}b^2 - 21^2}{2ab \cdot \frac{7}{20}} &= \frac{\frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2}{2ab \cdot \frac{7}{20} \cdot \frac{5}{3}} \\ \frac{5}{3} \cdot \left( a^2 + \frac{49}{400}b^2 - 21^2 \right) &= \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 \\ \frac53 a^2 + \frac{49}{240}b^2 - 21 \cdot 35 &= \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 \\ \frac{98}{1200} b^2 + 35 \cdot 14 &= \frac{10}{9} a^2 \\ \frac{49}{600} b^2 + 35 \cdot 14 &= \frac{40}{36} a^2. \end{align*} By using the Law of Cosines on $\triangle PAQ$ and $\triangle QAC$, we have \begin{align*} \frac{\frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2}{2ab \cdot \frac{7}{20} \cdot \frac{5}{3}} &= \frac{b^2 + \frac{25}{9}a^2 - 100^2}{2ab \cdot \frac{5}{3}} \\ \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 &= \frac{7}{20} \cdot \left( b^2 + \frac{25}{9}a^2 - 100^2 \right) \\ \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 &= \frac{7}{20} b^2 + \frac{35}{36} a^2 - 100 \cdot 35 \\ \frac{65}{36} a^2 + 65 \cdot 35 &= \frac{91}{400}b^2 \\ \frac{5}{36} a^2 + 5 \cdot 35 &= \frac{7}{400}b^2. \end{align*} Multiplying the second equation by $-8$ and adding the two equations results in \begin{align*} -\frac{40}{36} a^2 - 40 \cdot 35 &= -\frac{7}{50} b^2 \\ \frac{40}{36} a^2 &= \frac{49}{600} b^2 + 35 \cdot 14 \\ -40 \cdot 35 &= -\frac{35}{600} b^2 + 35 \cdot 14 \\ -40 &= -\frac{1}{600} b^2 + 14 \\ \frac{b^2}{600} &= 54 \\ b^2 &= 600 \cdot 9 \cdot 6 \\ b &= 6 \cdot 10 \cdot 3 \\ &= 180. \end{align*} After substituting $b$ back, solve for $a$ to get \begin{align*} \frac{49}{600} \cdot 180 \cdot 180 + 35 \cdot 14 &= \frac{40}{36} a^2 \\ 49 \cdot 54 + 35 \cdot 14 &= \frac{40}{36} a^2 \\ 49 \cdot 54 + 49 \cdot 10 &= \frac{40}{36} a^2 \\ 49 \cdot 64 &= \frac{10}{9} a^2 \\ a^2 &= \frac{49 \cdot 9 \cdot 64}{10} \\ a &= \frac{168}{\sqrt{10}} \\ &= \frac{84\sqrt{10}}{5} \end{align*} Thus, $\frac{AB}{AC} = \frac{84\sqrt{10}}{5} \cdot \frac{1}{180} = \frac{7\sqrt{10}}{75}$, so $p+q+r = \boxed{92}$.

See Also

2006 iTest (Problems)
Preceded by:
Problem 31
Followed by:
Problem 33
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