Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (Faster and less Calculation))
m (→‎Solution 2 (Faster and less Calculation))
 
Line 66: Line 66:
 
~CC2010CC2015
 
~CC2010CC2015
  
−
NOTE: NO TRIG REQUIRED< BTW ANOTHER WAY IS LAW OF SINES
+
NOTE: NO TRIG REQUIRED BTW ANOTHER WAY IS LAW OF SINES
  
 
==See Also==
 
==See Also==

Latest revision as of 21:49, 29 July 2025

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 $p+q+r = \boxed{92}$.

NOTE: SIMPLY USE STEWARTS THEOREM


Solution 2 (Faster and less Calculation)

Draw line $\overline{AD}$ where it is a bisector of $\angle PAQ$. Let $\overline{PD}$ be x. Then you can use angle bisctors $\overline{AP}$, $\overline{AD}$, and $\overline{AQ}$ to get the following ratios: \[\frac{\overline{AB}}{\overline{AQ}} = \frac{3}{5}\] \[\frac{\overline{AP}}{\overline{AC}} = \frac{7}{20}\] \[\frac{\overline{AQ}}{\overline{AP}} = \frac{35-x}{x}\]

Multiplying the ratios, we can find that $\frac{\overline{AB}}{\overline{AC}} = \frac{21(35-x)}{100x}$. On the other hand, using $\overline{AD}$ once again on the big triangle $\triangle{ABC}$, we find that $\frac{\overline{AB}}{\overline{AC}}$ also equals to $\frac{21+x}{135-x}$. If we try to calculate this directly, the numbers would be astronomically high. Therefore, we can introduce k. Let k be $\frac{\overline{AB}}{\overline{AC}}$. Then, the substitution for x in terms of k: $\frac{135k-21}{k+1}$. Plugging k in, we get: \[k = \frac {21(35-\frac{135k-21}{k+1})} {\frac{100(135k-21)}{k+1}}\]

Simplyfing, we cancel out the $2100$k's, and we get \[13500k^2 = 21\cdot56\] Solving for k, we get $k = \frac{7\sqrt{10}}{75}$

Then adding $7$, $10$, and $75$, we get $\boxed {92}$, as desired.

~CC2010CC2015

NOTE: NO TRIG REQUIRED BTW ANOTHER WAY IS LAW OF SINES

See Also

2006 iTest (Problems, Answer Key)
Preceded by:
Problem 31 		Followed by:
Problem 33
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_iTest_Problems/Problem_32&oldid=255960"