An Equilateral Triangle Inside a Square
This is an article on an olympiad problem. Here we present various solutions of the problem. We show the beauty of this problem by presenting different proofs to the same problem.
The Problem Statement: –
is a square.
is a point inside the square such that
. Show that
is equilateral.
A Solution by the use of Trigonometry
Construct the perpendicular bisector of and
and call it
. Since
is isosceles
passes through
. Call the length of the side of the square as
. Now in right angled triangled
use
and the tan of
to get the length
in terms of
. Now
. Hence use this and
to get
and hence
which we find equal to
. Similarly we can proceed for
and hence we are done.
First Synthetic Solution by Construction
We use proof by contradiction. The idea is that if what is given in the question is right then are all radii of the same circle with radius equal to the length of the side of the square and with center
. Let us assume that the point
does not fall on the circle. Then the line BE should meet the circle in some other point as it is already meeting it at the point
non-perpendicularly. Let that point be
. Our aim is to show that
.
In ,
(radii of the same circle). Hence it gives us
. Hence it leads to
. Similarly we get that
is isosceles giving us
equilateral. USing the same construction as in the first solution and using the fact that
is equilateral we get that
lies on
but
also lies on
. Hence,
. But
also lies on
and therefore
. Since two non-parallel straight lines can meet at only one point point hence
, a contradiction. Hence,
is equilateral.
A Solution by Subtracting
We construct equilateral inside the square and show that
.
A Solution by Using Inequalities
We suppose ,
,
,
. Then
Similarly we can show contradiction for the case when . Hence we have contradiction in both the cases. Hence,
.
Another Synthetic Solution
We erect on
to the interior. We join
. Now it is easy to see that
.
Yet Another Synthetic Solution
We erect regular on
to the exterior. Then
are isosceles, i.e.,
. Since,
giving us
.
Some remarks
The last 4 solutions are there in the books given in the references. The first was 3 are the ones I was able to come up on my own. Prof B.J. Venkatachala mentions here that this problem is one of his favourite problems
Acknowledgements
I would like to thank Prof. B. R. Sharma of Dibrugarh University for introducing me to this beautiful problem.
References
[1] M R Modak, S A Katre, V V Acharya, An Excursion in Mathematics, 7th edition, Bhaskaracharya Pratishthana, 2007.
[2] H S M Coxeter, Introduction to Geometry, 2nd edition, John Wiley & Sons, 1969.
[3] Arthur Engel, Problem-Solving Strategies, Springer, 1999.
[4] H S M Coxeter and S L Greitzer, Geometry Revisited, New Mathematical Library 19, The Mathematical Association of America, 1967.