ratio chasing inside a triangle, segment trisecting

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ratio chasing inside a triangle, segment trisecting

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Source: CRMO 2012 Region 2 p5

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#1 h Sep 30, 2018, 4:59 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle. Let $D, E$ be a points on the segment $BC$ such that $BD =DE = EC$ . Let $F$ be the mid-point of $AC$ . Let $BF$ intersect $AD$ in $P$ and $AE$ in $Q$ respectively. Determine $BP:PQ$ .

This post has been edited 1 time. Last edited by parmenides51, Sep 30, 2018, 6:38 PM
Reason: corrected , in the last sentence there was a typo, : instead of =

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Arkmmq

#2 h Sep 30, 2018, 5:09 PM • 1 Y

Y by Adventure10

Straightforward with barycentric coordinates

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#3 h Sep 30, 2018, 6:18 PM • 2 Y

Y by Adventure10, Mango247

here is an error, in conclusion,
the right thing is ΒΡ=ΡF

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#5 h Sep 30, 2018, 6:38 PM • 2 Y

Y by Adventure10, Mango247

yes there was a typo, the ratio of the lengths is wanted and not their equality

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#6 h Oct 1, 2018, 1:54 AM • 2 Y

Y by Adventure10, Mango247

Is the answer $4$ .

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#7 h Oct 1, 2018, 4:46 AM • 1 Y

Y by Adventure10

It is easy. Just using Menelaus Theorem many times we would find the result

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#8 h Oct 1, 2018, 11:52 AM • 2 Y

Y by Adventure10, Mango247

my solution is right, just made an accounting error, 1/3 instead of rectal 2/3:
$\begin{array}{l} \frac{{{\rm{BP}}}}{{{\rm{PQ}}}} = \frac{{{\rm{PF}}}}{{{\rm{PQ}}}} = \frac{{{\rm{PQ + QF}}}}{{{\rm{PQ}}}} = 1 + \frac{{{\rm{QF}}}}{{{\rm{QP}}}}\mathop {}\limits_{}^{} {\rm{ }}(1)\\ \frac{{{\rm{QF}}}}{{{\rm{PQ}}}} = \frac{{{\rm M}{\rm N}}}{{{\rm{P}}{\rm Z}}} = \frac{{\frac{{{\rm{DE}}}}{2}}}{{\frac{{{\rm M}{\rm N} + {\rm{DE}}}}{2}}} = \frac{2}{3}\mathop {}\limits_{}^{} \mathop \Rightarrow \limits^{(1)} \mathop {}\limits_{}^{} \frac{{{\rm{BP}}}}{{{\rm{PQ}}}} = \frac{5}{3} \end{array}$

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#9 h Oct 30, 2018, 9:22 AM • 3 Y

Y by RishiNandha_M, Adventure10, Mango247

I guess the correct answer for $\boxed {BP:PQ=5:3}$ .......that's what the official solution atleast says......

Ok so here's the solution using Barycentric Coordinates,

It is trivial to find out that $P\equiv (x:2:1)$ and since, $P$ lies on $BF$ , where $F \equiv (1:0:1)$ it follows that,
$\boxed {P \equiv (1:2:1)}$
Similarly, we get that $Q \equiv (2:y:2)$ and $Q \equiv (x:1:2)$
$\implies \boxed{Q \equiv (2:1:2)}$
Let WLOG, assume that $[ABC]=1$ , then from Barycentric Coordinates for $P$ and $Q$ , we have,

$[ABP]=\frac {1}{4} \text { and} [ABQ]=[ABP]+[APQ] = \frac {2}{5}$ Hence, it's easy to notice that $\frac {[ABQ]}{[ABP]} = \frac {[APQ]}{[ABP]} +1 = \frac {8}{5} \implies \frac {[APQ]}{[ABP]} =\frac {PQ}{BP} = \frac {3}{5}$
Hence, $\boxed {BP:PQ=5:3}$

This post has been edited 2 times. Last edited by AlastorMoody, Oct 31, 2018, 7:41 AM

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#10 h Oct 30, 2018, 12:50 PM • 2 Y

Y by Adventure10, Mango247

Menelaos in $\triangle ADC$ with transversal $\overline {BPF}$ gives $AP=3PD, PD$ is midline of $\triangle BEF\implies EF=2PD$ , so $\frac{AP}{EF}=\frac{3}2=\frac{PQ}{QF}$ , last relation being equivalent to $\frac{PQ}{PF}=\frac{3}5$ , but $PF=BP$ , hence $\frac{BP}{PQ}=\frac{5}3$ .

Best regards,
sunken rock

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bal09

#11 h May 31, 2024, 7:48 AM

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easy solution using midpoint theorm and similar traingles but for that we need to do construction i made attachment once check it for solution

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#12 h Mar 19, 2025, 1:40 PM

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i will bash.

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