Points X, Y are on sides CA and AB of triangle ABC and BX, CY meet at P. If AX:XC = BY:YA =1:2, find ratio BP:PX

Question

Points X, Y are taken on the sides CA, AB of triangle ABC if BX, CY meet at P and AX/XC =BY/YA $=1/2$, find the value of the ratio BP/PX.

Construction: Joining AP and extending it to cut BC at Z. Applying Ceva's theorem, BZ/ZC.CX/XA.AY/YB$=1$. Therefore, BZ/ZC$=1/4$

How do I proceed from here?

Answer 1

I'm sure that there are other, more elegant solutions, but here is a quick one, based on physics:

Let's assume that we put some weights in the corners of the triangle $\triangle ABC$, such that $m_C=1$, $m_A=2$, and $m_B=4$. Then the center of gravity between $A$ and $B$ is at $Y$, such as $$m_A\vec r_A+m_B\vec r_B=(m_A+m_B)\vec r_Y$$ You can see now that $$\frac{|\vec r_Y-\vec r_B|}{|\vec r_A-\vec r_Y|}=\frac{|m_A\vec r_A+m_B\vec r_B-m_A\vec r_B-m_B\vec r_B|}{|m_A\vec r_A+m_B\vec r_A-m_A\vec r_A-m_B\vec r_B|}=\frac{m_A}{m_B}=\frac24=\frac12$$

Similarly, you have $X$ as the center of gravity between $A$ and $C$. $$m_A\vec r_A+m_C\vec r_C=(m_A+m_C)\vec r_X$$Then $P$ is the center of gravity of the entire $ABC$ triangle. $$\vec r_P=\frac{m_A\vec r_A+m_B\vec r_B+m_C\vec r_C}{m_A+m_B+m_C}$$

Then $$\begin{align}\frac{BP}{PX}&=\frac{\vec r_P-\vec r_B}{\vec r_X-\vec r_P}\\&=\frac{m_A\vec r_A+m_B\vec r_B+m_C\vec r_C-(m_A+m_B+m_C)\vec r_B}{(m_A+m_B+m_C)\vec r_X-(m_A\vec r_A+m_B\vec r_B+m_C\vec r_C)}\\&=\frac{[m_A(\vec r_A-\vec r_B)+m_C(\vec r_C-\vec r_B)](m_A+m_C)}{(m_A+m_B+m_C)(m_A\vec r_A+m_C\vec r_C)-(m_A\vec r_A+m_B\vec r_B+m_C\vec r_C)(m_A+m_C)}\\&=\frac{[m_A(\vec r_A-\vec r_B)+m_C(\vec r_C-\vec r_B)](m_A+m_C)}{[m_A(\vec r_A-\vec r_B)+m_C(\vec r_C-\vec r_B)]m_B}\\&=\frac{m_A+m_C}{m_B}\\&=\frac 34\end{align}$$

I could have simplified the calculations if I would have chosen my origin to be $B$.

Answer 2

Let [.] denote areas. Then, relate the ratio to those of areas to get,

$$\frac{BP}{PX}=\frac{[YBC]}{[YXC]}=\frac{\frac13[ABC]}{\frac23[YAC]}=\frac{\frac13[ABC]}{\frac23\cdot \frac23[ABC]}=\frac34$$

Edit: As pointed out by @Andrei below, the first equality could be seen from drawing the heights of the two triangles and use similarity argument.

Answer 3

Draw median BT, draw YD parallel with AC, it crosses YD at Q so YQ=QD. Draw a line from D to X, , so AX=YD , so $QY=\frac{AX}{2}$

⇒ $QD=\frac{AX}{2}=\frac{1}{4}CX$ ⇒ $QD=TX=\frac{CT}{3}$

Now in triangle BXC we have:

$\frac{BZ}{ZC}\times\frac{CT}{TX}\times \frac{XP}{PB} =\frac{1}{4}\times \frac{3}{1}\times \frac{XP}{PB}=1$

⇒ $\frac{BP}{PX}=\frac{3}{4}$