How Many Integer Values Can QR Take in Triangle PQR with Equal Area Divisions?

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SUMMARY

The discussion focuses on determining the number of possible integer values for the length of side QR in triangle PQR, where PQ measures 7 units. The angle bisector QPR intersects line QR at point A, and points B and C on sides PR and QR, respectively, create parallel lines PA and BC that divide triangle PQR into three equal-area sections. The solutions provided by members laura123 and johng confirm the mathematical approach to solving this problem.

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Let $PQR$ be a triangle such that $PQ=7$ and let the angle bisector $QPR$ intersect line $QR$ at $A$. If there exist points $B$ and $C$ on sides $PR$ and $QR$ respectively, such that lines $PA$ and $BC$ are parallel and divide triangle $PQR$ into three parts of equal area, determine the number of possible integer values for $QR$.

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Hello MHB Community,

anemone is a bit under the weather this week, so she has asked me to fill in for her. Please join me in wishing for her a speedy recovery. (Yes)

Congratulations to the following members for their correct solutions:

  1. laura123
  2. johng

Solution from laura123:

The points $B$ and $C$ (such that lines $PA$ and $BC$ are parallel and divide triangle $PQR$ into three parts of equal area) there exist if the area of the triangle $PRA$ is equal to twice the area of the triangle $PAQ$.
Let $A_1$ be the area of the triangle $PAQ$ and $A_2$ the area of the triangle $PRA$.
$A_2=2A_1$ if $PR=2PQ$ in fact:
$A_1=\dfrac{1}{2}PQ\cdot PA\cdot \sin(\angle QPA)$;
$A_2=\dfrac{1}{2}PR\cdot PA\cdot \sin(\angle APR)$.
Since $\angle QPA=\angle APR$ ($PA$ is the bisector of $\angle QPR$) and $A_2=2A_1$ it follows $PR=2PQ$.
Since $PQ=7$, we have $PR=14$.
10dyvs2.png

Therefore, the vertex $R$ must belong to a circle with centre $P$ and radius $14$, as shown in the following figure:
2hmhrly.png

then: $7<QR<21$. Therefore, the possible integer values for $QR$ are:
8,9,10,11,12,13,14,15,16,17,18,19,20.
There are 13 possible integer values for $QR$.

Solution from johng:

2rms3mx.png


72fxnk.png
 

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