You may also like

Vedic Sutra - All from 9 and Last from 10

Vedic Sutra is one of many ancient Indian sutras which involves a cross subtraction method. Can you give a good explanation of WHY it works?

Tournament Scheduling

Scheduling games is a little more challenging than one might desire. Here are some tournament formats that sport schedulers use.

Archimedes and Numerical Roots

The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?

Triangle Incircle Iteration

Age 14 to 16 Challenge Level:

Meg offers this solution:

The tangent of a circle is at right-angles to the radius of the circle. That is, if you join the centre point of the circle to a point where the circle meets the outer triangle, it makes an angle of $ 90 ^{\circ} $ with the side of the triangle.

triangle with inscribed circle and radius
The bisector of the angles of triangle $ABC$ will all pass through the centre of the circle.
triangle with inscribed circle and bisectors
From this we know that
$OAZ = \frac {a}{2}$ and $OZA = 90 ^{\circ}$
Hence $AOZ = 90- \frac{a}{2}$
Now consider triangle XOZ. This triangle is isosceles, so $OXZ = XZO = \frac{180-(180-a)}{2} = \frac{a}{2}$

By similar arguments
$OXY = OYZ = \frac{b}{2}$ and $OYZ = OZY = \frac{c}{2}$ Hence the new angles of the triangle are

$ZXY= \frac{a}{2}+\frac{b}{2}$

$XYZ = \frac{b}{2} + \frac{c}{2}$

$YZX =\frac{c}{2} + \frac{a}{2} $

$a+b+c=180 $

Hence $ \frac{a+b}{2}= 90 - \frac{c}{2}$

It follows that $ZXY =90 - \frac {c}{2}$.

A similar argument can be followed for $XYZ$ and $YZX$. If you continue drawing triangles within circles, the angles will decrease as shown here:

Triangle 1: a
Triangle 2: $ 90 - \frac{a}{2}$
Triangle 3: $90- \frac{90-\frac{a}{2}}{2}$ =$90 - \frac{90}{2} + \frac{a}{4}$
Triangle 4: $90- \frac{90-\frac{90}{2}+\frac{a}{4}}{2}= \frac{3}{4}.90 - \frac{a}{8}$

When you continue this iteration, you can demonstrate that the $a$ term becomes less and less significant, and the sum of the rest of the terms tends to 60 degrees. Hence the triangle tends to an equilateral triangle.