The question states: a triangle ABC has acute angles at A and B, where |BC| = a, |CA| = b and |AB| = c. The point P lies on the side AB. Find an expression for |CP|^2.

An example of the triangle described

The correct answer is

$$\frac{a^2}{c}\cdot |AP|+\frac {b^2}c\cdot (c-|AP|)-|AP|\cdot |BP|$$

I have tried using the formula for the sides of a kite however it gets very messy and complicated. I suspect there is a more elegant solution. Here's my thought process on a diagram. Picture of triangle and my kite

  • $\begingroup$ Use the law of cosines in triangles $CPA$ and $CPB$ $\endgroup$ – Lozenges Apr 29 at 19:36

As @Lozenges suggests in the comments, whenever you have such a configuration, the Law of Cosines might help. Why? Easy: because $\color{blue}{\cos(\angle APC)=-\cos (\angle CPB)}$. This might not help a lot, but you'll see later how we can take advantage of this fact!

Now, in virtue of the Law of Cosines in $\triangle CPA$ and $\triangle CPB$, we obtain $$\begin{align*}a^2&=PC^2+PB^2-2PC\cdot PB\cdot \cos (\angle CPB)\\b^2&= PC^2+AP^2\color{blue}+2PC\cdot AP\cdot\color{blue}{\cos (\angle CPB)}\end{align*}$$ Or equivalently

$$\begin{align*}a^2\color{brown}{\cdot AP}&=PC^2\color{brown}{\cdot AP}+PB^2\color{brown}{\cdot AP}-2PC\cdot PB\color{brown}{\cdot AP}\cdot \cos (\angle CPB)\tag{1}\\b^2\color{brown}{\cdot PB}&= PC^2\color{brown}{\cdot PB}+AP^2\color{brown}{\cdot PB}+2PC\color{brown}{\cdot PB}\cdot AP\cdot\cos (\angle CPB)\tag{2}\end{align*}$$

Why doing this? Well, adding $(1)$ and $(2)$ yields (the cosines magically disappear!)

$$a^2\cdot AP+b^2\cdot PB=PC^2(AP+PB)+AP\cdot PB(AP+PB)=PC^2\cdot c+AP\cdot PB\cdot c$$

Do you recognize it?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.