You may also like

At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

Contact

A circular plate rolls in contact with the sides of a rectangular tray. How much of its circumference comes into contact with the sides of the tray when it rolls around one circuit?

Gift of Gems

Four jewellers share their stock. Can you work out the relative values of their gems?

Half a Triangle

Age 14 to 16 Challenge Level:

Well done Arun from National Public School, Bangalore, India, some quality thinking in devising this solution.

We are given a triangle $ABC$ , and are required to draw a line $DE$ parallel to $CB$ such that it divides the triangle into $2$ of equal areas.

The area of triangle $ABC$ is double the area of $AED$.

But, $ADE$ and $ACB$ are similar triangles because $DE$ is parallel to $CB$

We also know that the ratio of the areas of the two similar triangles is equal to the ratio of the squares of corresponding sides.

Which means that the line ratio $AD$:$AC$ must be $1 : \sqrt{2}$
The problem becomes : how to locate $D$ to achieve this ratio.

A square of side length $1$ has a diagonal length of$ \sqrt{2}$

or, put another way, an isosceles right-angled triangle has a hypotenuse $ \sqrt{2}$ times bigger than the other sides.

Here is a construction to achieve this required ratio.

$X$ is any suitable point on $AD$

$ZX$ is perpendicular to $AC$, and $ZX$ is equal in length to $AX$.

So $AXZ$ is an isosceles right-angled triangle.

By sweeping an arc centre $A$ from $X$ to $AZ$ at $N$, $AN$ is made equal to $AX$

$AN$ to $AZ$ is now in the required ratio.

Drawing from $N$ parallel to $ZC$ the point $D$ is reached.

Because $AND$ and $AZC$ are similar triangles, $AD$ and $AC$ are in the required ratio.

Excellent and simple!