Area Problem 2

We use the three midpoints of the sides of $\triangle ABC$ , plus one of its vertices, to form a parallelogram; then we use this parallelogram to tile the plane, as shown below. The image on the left is of $\triangle ABC$ ; the one on the right shows $\triangle PQR$ , formed from the medians of $\triangle ABC$ .

If we use the parallelogram in the tiling as a unit of area, then clearly the area of $\triangle ABC$ is two such units. $\triangle PQR$ 's area is not as clear; however, it's easy to see that $\triangle PQR$ has the same area as $\triangle QRS$ (because $\triangle PQT \cong \triangle RST$ ), whose area is clearly one and a half units.

Thus the area of the triangle formed by the medians of $\triangle ABC$ is $\dfrac{3}{4}$ of the area of $\triangle ABC$ , i.e. $\dfrac{3}{4}(120) = \boxed{90}$

Insight: Use vector and cross product.

Some useful properties:

$\text{1. } \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}$

$\text{2. } \vec{a} \times \vec{a} = \vec{0}$

$\text{3. in } \Delta ABC , \vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$

$\text{4. area of } \Delta ABC = \frac{1}{2} | \vec{AB} \times \vec{AC} |$

To make things easier to read, $\text{let } \vec{BC} = \vec{a} , \vec{CA} = \vec{b} , \vec{AB} = \vec{c}$

$\text{1. } \vec{a} + \vec{b} + \vec{c} = \vec{0}$

$\text{2. } \vec{AD} = \vec{c} + \frac{1}{2} \vec{a} , \quad \vec{BE} = \vec{a} + \frac{1}{2} \vec{b} , \quad \vec{CF} = \vec{b} + \frac{1}{2} \vec{c}$

$\text{3. } \vec{AD} + \vec{BE} + \vec{CF} = \frac{3}{2} ( \vec{a} + \vec{b} + \vec{c} ) = \vec{0} \implies \vec{AD} , \vec{BE} , \vec{CF} \text{ form a triangle which is congruent to } \Delta PQR.$

$\text{4. let } | \vec{S} | = \text{area of } \Delta ABC , 2\vec{S} = \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$

$\text{5. area of } \Delta PQR$

$= | \frac{1}{2} \vec{AD} \times \vec{BE} |$

$= | \frac{1}{2} ( \vec{c} + \frac{1}{2} \vec{a} ) \times ( \vec{a} + \frac{1}{2} \vec{b} ) |$

$= | \frac{1}{2} \vec{c} \times \vec{a} + \frac{1}{4} \vec{c} \times \vec{b} + \frac{1}{4} \vec{a} \times \vec{a} + \frac{1}{8} \vec{a} \times \vec{b} |$

$= | \frac{1}{2} \vec{c} \times \vec{a} - \frac{1}{4} \vec{b} \times \vec{c} + \frac{1}{4} \vec{0} + \frac{1}{8} \vec{a} \times \vec{b} |$

$= | \vec{S} - \frac{1}{2} \vec{S} + \frac{1}{4} \vec{S}|$

$= | \frac{3}{4} \vec{S}|$

$= 90$

The three medians meet at a centroid $M$ , which by centroid properties means that the centroid cuts each median in a $2:1$ ratio, and the medians partition the triangle into $6$ congruent triangles (see here ). Let $\triangle A'B'C' \cong \triangle ABC$ , and rotate $\triangle A'B'C$ so that parallelogram $A'BAC$ is formed, as shown below.

Let $A_o$ be the area of the original triangle $\triangle ABC$ , $A_m$ be the area of the triangle formed by the lengths of the medians, and $A_n$ be the area of \triangle $MCM'$ . Since $\triangle MCM'$ is made up of two congruent triangles that are $\frac{1}{6}$ of the original triangle's area, $A_n = \frac{1}{3} A_o$ . Also, the sides of $\triangle MCM'$ are each equal to $\frac{2}{3}$ of the medians of $\triangle ABC$ , so there is a $\frac{2^2}{3^2} = \frac{4}{9}$ area ratio, which means $A_n = \frac{4}{9}A_m$ . Since $A_n = \frac{1}{3} A_o = \frac{4}{9}A_m$ , $A_m = \frac{3}{4}A_o$ .

Therefore, if the area of the original triangle $A_o = 120$ , the area of the triangle formed by the lengths of the medians $A_m = \frac{3}{4} \cdot 120 = \boxed{90}$ .

Since there is no constraint on the type of triangle ABC is we can always assume it to be equilateral triangle with a side length 'a' and still get the right result and also simplify our solutions.