# Geometry puzzle [closed] The lengths of the triangle sides satisfy $x < y < z$. Lines that look parallel indeed are parallel. Triangles that look similar indeed are similar. The three red lines are concurrent (meet at a single point) and congruent (have the same length). What is their length?

• Do you want their length in terms of x, y, and z? – Joe Z. Mar 17 '16 at 16:30
• Yeah, a general solution in terms of x, y, and z. I saw this puzzle posed specifically for a 3:4:6 triangle, and hammered away at it for a while before going to bed. – Anon Mar 17 '16 at 16:42
• Nice solutions... So the answer to the 3:4:6 triangle is 8. – Gamow Mar 17 '16 at 18:11

Let $a$ be the proportion of the dimensions of the triangle including the line from $y$ to $z$ to those of the full triangle. Define $b$ and $c$ similarly for the lines from $x$ to $z$, and from $x$ to $y$. Then, from similarity, $ax=by=cz$.

Also, if we look at side $x$, we can see from similarity that the section from the bottom to the $xz$ line has length $bx$, the section from the top to the $xy$ line has length $cx$, and the overlap of those two sections (the section between the two red lines) has length $(1-a)x$. Thus

$bx + cx - (1-a)x = x$

$\implies a+b+c=2$

Substituting $b=\frac{ax}{y}$ and $c=\frac{ax}{z}$, we have

$a+\frac{ax}{y}+\frac{ax}{z}=2$

$\implies a(\frac{yz+xz+xy}{yz})=2$

$\implies ax=\frac{2xyz}{xy+yz+zx}$

Since the three lines are congruent, this is the length of all of them.

• Duuude...spoilers? :) – Marius Mar 17 '16 at 17:58

$\frac{2*x*y*z}{x*y + x*z + y*z}$

Proof: Let's convene to the notations on the image:

And let's make the length of the red lines $q$. $AxH = AyA$
$BxH = BzB$
This means $AB = AxBx + AyBz$ which translates to $AyBz = x-q$
In a similar logic $AxCz = y-q$
Triangle AyBzH is similar to ABC This results in $\frac{AyBz}{x} = \frac{BzH}{z}$
Triangle AxCzH is similar to ABC This results in $\frac{AxCz}{y} = \frac{CzH}{z}$
Adding the 2 above and replacing $AxBz$ with $y-q$ and $AyBz$ with $x-q$ we end up with
$\frac{x-q}{x} + \frac{y-q}{y} = \frac{BzH}{z} + \frac{CzH}{z}$
This is the same as
$1- \frac{q}{x} + 1 - \frac{q}{y} = \frac{BzH + CzH}{z}$
But $BzH +CzH = q$
So $1- \frac{q}{x} + 1 - \frac{q}{y} = \frac{q}{z}$
Resolving the equation for unknown $q$ we get
$q = \frac{2*x*y*z}{x*y + x*z + y*z}$