Triangle in Triangle

Question

I have the lengths of three sides of an acute triangle ABC as shown below. Assume a point P on the side AB such that, if Q is the projection of P onto BC, R is the projection of Q onto CA, P becomes the projection of R onto AB. How can I Find the length PB.

Answer 1

I start with coordinates, choosing one edge length equal to $1$ w.l.o.g.:

$$B=(0,0) \qquad A=(x,y) \qquad C=(1,0) \qquad P=\lambda A+(1-\lambda)B$$

with $y\neq 0$ and some $\lambda\in[0,1]$ we'll need to determine. Then using some computation (which I did using projective geometry, but there are other ways as well) you get

$$ R=(x^3\lambda + xy^2\lambda - x^2\lambda - y^2\lambda + y^2, y (x^2\lambda + y^2\lambda - x))/(x^2 + y^2 - x) \\ Q=(x^4\lambda + 2x^2y^2\lambda + y^4\lambda - 2x^3\lambda - 2xy^2\lambda + x^2\lambda + y^2\lambda - y^2,0)/((x - 1)(x^2 + y^2 - x)) $$

The line $QP$ is then orthogonal to $BC$ iff they have the same $x$ coordinate, i.e. if

$$x^4\lambda + 2x^2y^2\lambda + y^4\lambda - 2x^3\lambda - 2xy^2\lambda + x^2\lambda + y^2\lambda - y^2 = \lambda x(x - 1)(x^2 + y^2 - x)$$

Subtract the right hand side, cancel out a common factor of $y^2$ and you end up with

$$\lambda=\frac1{x^2 + y^2 - x + 1}$$

Now let's turn that back into triangle lengths. I had edge lengths

$$a^2 = BC^2 = 1 \qquad c^2 = AB^2 = x^2+y^2 \qquad b^2 = AC^2 = (1-x)^2+y^2$$

so I had

$$a^2+b^2+c^2 = 2x^2+2y^2-2x+2$$

and therefore

$$\lambda=\frac{2a^2}{a^2+b^2+c^2}$$

Note that this formulation of the result is invariant under rescaling: if you scale all the coordinates by the same amount, then that amount cancels out of the above expression. Which fits in nicely with the fact that the convex combination of $P$ describes the same point no matter the scale. So at this point we are justified to drop the assumption of $a=1$ and are therefore back in the general case.

But you were not asking about this factor $\lambda$ but instead about the length $PB$. For this you have to multiply by $AB=c$, so the final result is

$$PB=\frac{2a^2c}{a^2+b^2+c^2}$$

Answer 2

Here is another approach, which works directly and systematically. Call the angles of the triangle $A, B, C$ and the sides opposite those angles $a, b, c$. We use the cosine rule $a^2=b^2+c^2-2bc \cos A$ so that $\cos A=\cfrac {a^2+b^2+c^2-2a^2}{2bc}$ and if I define $2d^2=a^2+b^2+c^2$ then $\cos A=\cfrac {d^2-a^2}{bc}, \cos B=\cfrac {d^2-b^2}{ac}, \cos C=\cfrac {d^2-c^2}{ab}$.

Let BP=p.

Then $$p=c-AP=c-AR\cos A= c-(b-CR)\cos A=c-b\cos A+CR\cos A=$$

$$=c-b\cos A+CQ\cos C\cos A=c-b\cos A+(a-BQ)\cos C\cos A=$$

$$=c-b\cos A+a\cos A\cos C-BQ\cos A\cos C=$$

$$=c-b\cos A+a\cos A\cos C-p\cos A\cos B\cos C$$

So that $$p\left(1+\cos A\cos B\cos C\right)=c-b\cos A+a\cos A\cos C=$$$$=c-\frac {d^2-a^2}c+\frac {(d^2-a^2)(d^2-c^2)}{b^2c}=\frac {b^2c^2-b^2d^2+b^2a^2+d^4-a^2d^2-c^2d^2+a^2c^2}{b^2c}$$

Now with $a^2+b^2+c^2=2d^2$ this is equal to $$\frac {a^2b^2+b^2c^2+a^2c^2-d^4}{b^2c}$$

We also have $$1+\cos A\cos B\cos C=\frac {a^2b^2c^2+d^6-(a^2+b^2+c^2)d^4+(a^2b^2+b^2c^2+a^2c^2)d^2-a^2b^2c^2}{a^2b^2c^2}=\frac {d^2(a^2b^2+b^2c^2+c^2a^2-d^4)}{a^2b^2c^2}$$

So that everything cancels nicely to give $$p=\frac {a^2c}{d^2}=\frac {2a^2c}{a^2+b^2+c^2}$$