A Partitioned Equilateral Triangle

Geometry Level 3

Above shows an equilateral triangle $ABC$ with side length 1. Given that $M, N$ and $Q$ are points on the sides $AB, BC$ and $AC$ such that the lines $AN, BQ$ and $CM$ divide the triangle into 4 triangles and 3 quadrilaterals.

We color the triangles in two colors (orange and green) in such way that any two triangle with a common vertex are colored with different colors.

If the area colored in green is equal to that colored in orange, find the value of $AM+BN+CQ$ .

The answer is 1.

1 solution

Chew-Seong Cheong
Oct 28, 2018

Let $AM$ , $BN$ and $CQ$ be $x$ , $y$ and $z$ respectively. We need to find $x+y+z$ , let the area of the equilateral triangle be $a$ .

Because the side length of equilateral $\triangle ABC$ is 1, the area of $\triangle ABN$ , $[ABN]=ya$ and that of green $\triangle APM$ , $[APM]=xya$ . Similarly, $[BNR]=yza$ and $[CQS]=zxa$ . Therefore, the area of the three green triangles is $(xy+yz+zx)a$ .

The area of the orange $\triangle PRS$ is given by:

$\begin{aligned} [PRS] & = [CBM] - {\color{#3D99F6}[PNBM]} - {\color{#D61F06}[CNRS]} \\ & = [CBM] - {\color{#3D99F6}([ABN]-[APM])} - {\color{#D61F06}([BCM]-[CQS]-[BNR])} \\ & = \big((1-x) - {\color{#3D99F6}(y-xy)} - {\color{#D61F06}(z-zx-yz)}\big)a \\ & = (1-x-y-z+xy+yz+zx)a \end{aligned}$

Since the area in green is equal to that in orange, we have:

$\begin{aligned} (xy+yz+zx) a & = (1-x-y-z+xy+yz+zx)a \\ 0 & = 1-x-y-z \\ \implies x+y + z & = \boxed 1 \end{aligned}$

A Partitioned Equilateral Triangle

The answer is 1.

1 solution

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