## 01 Sep 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.

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I am a masters student at Université Paris-Est . I am also an editor of the English section.

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