Geometry question with a triangle

[Dank2]

TL;DR Summary

Geometry question with triangle

AB=AC. P is on ac such that AP=3PC. Q on CB such that CQ=3BQ.
Need to find the length of PQ.

I know i can use the Cosine theorem, but the answer is without Cosine.

 

[BvU]

 

[mfb]

You can use the cosine theorem and then replace that cosine by a different expression.

 

[BvU]

I know i can use the Cosine theorem

post your work -- it's required to get assistance in PF

 

[Dank2]

Thanks, solved using cosine for the two triangles

 

[aheight]

I obtain a different answer but perhaps I'm not setting up the problem correctly. As I understand it, we have an isosceles triangle and the ratios are 3:1 or PB=1/4 of AB and CQ=1/4 of CB. Is this not correct? If so then I obtain $PQ=1/4\sqrt{(AB)^2+6(CB)^2}$. In the diagram below, the height is 10, the base is 5 with $AB=AC=\sqrt{10^2+(5/2)^2}$. In that case $PQ\approx 4.00195$. Might someone explain where I went wrong?

 

[mfb]

OP forgot to square a "b" in the line that starts with the equal sign, apart from that the result is the same.

 

[aheight]

Thanks for showing me that mfb.

Can I ask if the problem is solvable without invoking the cosine law strictly through geometric means?

 

[mfb]

With Pythagoras only:

Put B at the origin, BC to the right along the x direction, BA (length c) upwards/right. Let a be the length CB.

C=(0, 0)
B=(a, 0)
A=(a/2, $\sqrt{c^2-a^2/4}$)
Therefore:
Q=(3a/4, 0)
P=(a/8, 1/4$\sqrt{c^2-a^2/4}$)
Using Pythagoras again, $PQ = \sqrt{(5a/8)^2 + 1/4^2 (c^2-a^2/4)} = \frac{1}{4}\sqrt{6a^2 + c^2}$

I don't think you can avoid anything like that completely. Square roots typically mean you need Pythagoras or more.

 

[aheight]

With Pythagoras only:

Put B at the origin, BC to the right along the x direction, BA (length c) upwards/right. Let a be the length CB.

C=(0, 0)
B=(a, 0)
A=(a/2, $\sqrt{c^2-a^2/4}$)
Therefore:
Q=(3a/4, 0)
P=(a/8, 1/4$\sqrt{c^2-a^2/4}$)
Using Pythagoras again, $PQ = \sqrt{(5a/8)^2 + 1/4^2 (c^2-a^2/4)} = \frac{1}{4}\sqrt{6a^2 + c^2}$

I don't think you can avoid anything like that completely. Square roots typically mean you need Pythagoras or more.

Thanks mfb! Here's how I worked through your solution:

1. Basically, using similar purple and red triangles below, we need to determine $x$ and $y$ in the diagram below. So we have
$$
\begin{array}{c}
\frac{\sqrt{c^2-a^2/4}}{a/2}=y/x\\
(c/4)^2=x^2+y^2
\end{array}$$

Once we have $x$ and $y$ , we can then use $(QP)^2=(QR)^2+y^2$ although quite a bit of algebra to get from P to Q.