Find the value of the ratio of m/n

[anemone]

In triangle $PQR$, the points $A,\,B,\,C$ lie on the line segments $QR$, $RP$ and $PQ$ respectively, such that $A$ is the midpoint of $QR$, $RB=3BP$ and $\dfrac{PC}{CQ}=\dfrac{m}{n}$. If $y$ is the area of triangle $RAB$, $x$ is the area of triangle $AQC$ and $z$ is the area of triangle $PCB$, and $x^2=yz$, find the value of $\dfrac{m}{n}$.

 

[Saitama]

Re: Find the value of the ratio of p/q

In triangle $PQR$, the points $A,\,B,\,C$ lie on the line segments $QR$, $RP$ and $PQ$ respectively, such that $A$ is the midpoint of $QR$, $RB=3BP$ and $\dfrac{PC}{CQ}=\dfrac{m}{n}$. If $y$ is the area of triangle $RAB$, $x$ is the area of triangle $AQC$ and $z$ is the area of triangle $PCB$, and $x^2=yz$, find the value of $\dfrac{m}{n}$.

Spoiler

Consider the figure below.
View attachment 2560

Choose $Q$ as the origin and the positions of $P$ and $R$ are denoted by vectors $\vec{p}$ and $\vec{r}$.

Also, let $m/n=\lambda$.

Clearly,
$$\vec{QC}=\frac{1}{\lambda+1}\vec{p}$$
$$\vec{QA}=\frac{\vec{r}}{2}$$
$$\vec{QB}=\frac{\vec{r}+3\vec{p}}{4}$$

Next, I find the areas in terms of $\vec{r}$ and $\vec{p}$,
$$x=\frac{1}{2}\left| \vec{QA}\times \vec{QC}\right|=\frac{1}{4(\lambda+1)}\left|\vec{r}\times \vec{p}\right|$$
$$y=\frac{1}{2}\left|\vec{AR}\times \vec{AB}\right|=\frac{1}{2}\left|\vec{AR}\times \left(\vec{QB}-\vec{QA}\right)\right|=\frac{1}{2}\left|\frac{\vec{r}}{2}\times \left(\frac{\vec{r}+3\vec{p}}{4}-\frac{\vec{r}}{2}\right)\right|=\frac{3}{16}\left|\vec{r}\times \vec{p}\right|$$
$$z=\frac{1}{2}\left|\vec{CP}\times \vec{CB}\right|=\frac{1}{2}\left| \frac{\lambda}{\lambda+1}\vec{p}\times \left(\vec{QB}-\vec{QC}\right)\right|=\frac{\lambda}{8(\lambda+1)}\left|\vec{r}\times \vec{p}\right|$$
As per the question,
$$x^2=yz \Rightarrow \frac{1}{16(\lambda+1)^2}=\frac{3}{16}\cdot \frac{\lambda}{8(\lambda+1)}\Rightarrow 3\lambda^2+3\lambda-8=0$$
$$\Rightarrow \boxed{\lambda=\dfrac{1}{6}\left(\sqrt{105}-3\right)}$$
(neglecting the negative root)

 

[lfdahl]

My suggested solution:

Spoiler

Please refer to the attached figure.

View attachment 2569

\[PQ = m+n,\: \: QR = 2p \: \: and \: \: RP=4q\\\\ x = \frac{1}{2}\cdot n\cdot p\cdot sin(Q)\\\\ y = \frac{3}{2}\cdot q\cdot p\cdot sin(R)\\\\ z = \frac{1}{2}\cdot q\cdot m\cdot sin(P)\]
\[x^2 = yz \Rightarrow n^2\cdot p^2\cdot sin^2(Q)=3\cdot q^2\cdot p\cdot m\cdot sin(R)\cdot sin(P)\;\;\;\;(1).\]

The rule of sines:

\[\frac{2p}{sin(P)}=\frac{4q}{sin(Q)}=\frac{m+n}{sin(R)}\\\\ \Rightarrow p^2 = 4\cdot q^2\cdot \frac{sin^2(P)}{sin^2(Q)} \: \: \: \: and\: \:\: \: p = \frac{1}{2}(m+n) \frac{sin(P)}{sin(R)}\]

Insertion into $(1)$ yields:

\[n^2\cdot p^2 \cdot sin^2(Q) = 4\cdot n^2\cdot q^2\cdot \frac{sin^2(P)}{sin^2(Q)}\cdot sin^2(Q)=3\cdot q^2\cdot p\cdot m\cdot sin(R)\cdot sin(P)\\\\ \Rightarrow 4\cdot n^2\cdot sin(P)=3\cdot p\cdot m \cdot sin(R) =\frac{3}{2}\cdot m\cdot (m+n)\cdot \frac{sin(P)}{sin(R)}\cdot sin(R) \\\\ \Rightarrow \left ( \frac{m}{n} \right )^2+\frac{m}{n}-\frac{8}{3}=0 \Rightarrow \frac{m}{n}=\frac{1}{2}\left ( \sqrt{\frac{35}{3}}-1 \right )\approx 1.208\]

 

[anemone]

Thanks to both of you for participating in this problem!:)

It's always great to see there are different methods to tackle a problem, thank you again for your willingness to participate and typing your solution so well in $\LaTeX$ for me and the readers! (Clapping)

 

[CellCoree]

I would approach this problem by first analyzing the given information and identifying any relevant mathematical principles or equations that can help solve for the ratio of m/n. From the given information, it is clear that the triangle $PQR$ is divided into three smaller triangles, $RAB$, $AQC$, and $PCB$, by the points $A$, $B$, and $C$. Additionally, we know that $A$ is the midpoint of $QR$, $RB=3BP$, and $\dfrac{PC}{CQ}=\dfrac{m}{n}$.

To find the value of $\dfrac{m}{n}$, we can use the fact that the areas of similar triangles are proportional to the squares of their corresponding sides. This means that the ratio of the areas of $RAB$ and $AQC$ is equal to the ratio of the squares of their corresponding sides, which are $RA$ and $AC$. Similarly, the ratio of the areas of $AQC$ and $PCB$ is equal to the ratio of the squares of their corresponding sides, which are $AC$ and $CP$.

Using this information, we can write the following equations:

$\dfrac{y}{x}=\left(\dfrac{RA}{AC}\right)^2$ and $\dfrac{x}{z}=\left(\dfrac{AC}{CP}\right)^2$

Since we are given that $x^2=yz$, we can substitute this into the second equation, giving us:

$\dfrac{y}{x}=\dfrac{x}{z}\Rightarrow \dfrac{y}{x}=\dfrac{x^2}{x^2}\Rightarrow \dfrac{y}{x}=1$

Thus, we can conclude that $\left(\dfrac{RA}{AC}\right)^2=1$, which means that $RA=AC$. This also implies that $RB=3BP=2AC$, since $A$ is the midpoint of $QR$. From this, we can see that $RA=2BP$ and $AC=BP$.

Now, using the fact that $RB=3BP$, we can write the following equation:

$RB=3BP=2AC \Rightarrow 2BP=3BP \Rightarrow BP=2AC$

Substituting this into the equation $\dfrac{PC}{CQ}=\dfrac{m