A cute geometry problem
=======================
:slug: a-cute-geometry-problem
:date: 2010-03-11 19:57:13+00:00
:updated: 2010-05-06 23:26:10+00:00
:tags: musings, science & technology
:category: rumblings
:has_math: yes
I came across a cute geometry problem recently and I would like to pass it
along.
Problem
-------
Problem Statement
~~~~~~~~~~~~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
If the sides of the square are of unit length and all curves are
circular arcs, what is the area of the highlighted region?
Although substantially easier with the use of calculus or trigonometry,
this problem can be solved entirely with basic geometry (no weird laws
you might have forgotten since high school are necessary).
I have derived a geometric solution, which follows, but I highly
recommend trying to do it yourself first.
.. TEASER_END
Step 1
~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
The highlighted section is the full square minus a quarter circle with
radius of unit length. The highlighted area can be found by simple
subtraction.
.. math::
A = 1 - \frac{\pi}{4}
Step 2
~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
Since the edges of the highlighted triangle are all on circular arcs
of unit length, the highlighted triangle is an equilateral triangle.
That the triangle is equilateral means that the intersection points
between the arcs trisect each arc. That the triangle is equilateral
and has unit length sides further allows us to calculate the height of
the triangle.
.. math::
h = \frac{\sqrt{3}}{2}
Step 3
~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
Having noted in Step 2 that the arc intersections trisect the arcs,
the highlighted wedge becomes one twelfth of a circle with radius of
unit length. The area can be found simply.
.. math::
A = \frac{\pi}{12}
Step 4
~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
Since we know the height of the triangle from Step 2, we can subtract
that from the height of the total square to get the left side of the
highlighted trapezoid. The right side of the highlighted trapezoid is,
of course, unit length. The width of the trapezoid, by symmetry, is one
half unit length. Knowing both sides and the width, we can calculate
the area of the highlighted trapezoid.
.. math::
A = \frac{1}{2} \cdot (1 + (1 - \frac{\sqrt{3}}{2})) \cdot \frac{1}{2} = \frac{4 - \sqrt{3}}{8}
Step 5
~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
By subtracting the area found in Step 3 from the area found in Step 4,
we can find the highlighted area.
.. math::
A = \frac{4 - \sqrt{3}}{8} - \frac{\pi}{12} = \frac{12 - 3 \sqrt{3} - 2 \pi}{24}
Step 6
~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
By symmetry, we know that the highlighted area is twice the area found
in Step 5.
.. math::
A = \frac{12 - 3 \sqrt{3} - 2 \pi}{12}
Step 7
~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
By subtracting the area found in Step 6 from the area found in Step 1,
we can find the highlighted area.
.. math::
A = (1 - \frac{\pi}{4}) - \frac{12 - 3 \sqrt{3} - 2 \pi}{12} = \frac{3 \sqrt{3} - \pi}{12}
Step 8
~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
By symmetry, we know that the highlighted area is four times the area
found in Step 7.
.. math::
A = \frac{3 \sqrt{3} - \pi}{3} = \sqrt{3} - \frac{\pi}{3}
Solution
~~~~~~~~
.. container:: row
.. container:: four columns
.. raw:: html
.. container:: eight columns
By subtracting the area found in Step 8 from the area of a unit length
sided square, we can obtain the area of the highlighted region.
.. math::
A = 1 - (\sqrt{3} - \frac{\pi}{3}) = 1 + \frac{\pi}{3} - \sqrt{3} \approx 0.3151467