  How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Excursion in Geometry - INMO 2019 Problem 1

Sequential hints for geometry problems are useful. It is much better than 'reading solutions'. Use this interesting problem from Indian National Math Olympiad (INMO 2019) to refresh your geometry skills.

Angle Chasing can be one of the most used tools in geometry. But it nevertheless requires some additional insight to finish off the problem.

## Problem

Let $ABC$ be a triangle with $\angle BAC > 90^\circ$. Let $D$ be a point on the line segment $BC$ and $E$ be a point on the line $AD$ such that $AB$ is tangent to the circumcircle of triangle $ACD$ at $A$ and $BE$ perpendicular to $AD$. Given that $CA=CD$ and $AE=CE$, determine $\angle BCA$ in degrees

Also see

## Hint 2: Alternate Segment Theorem

There is a relation between tangent and chord (and the angles they create). Here you go:

Crux of the theorem: Suppose BA is tangent to a circle at the point A. From A, if we draw a chord, then two angles are created:

1. $\angle BAD$ = angle between the tangent and chord
2. $\angle DCA$ = angle subtended (created) by the chord DA at any point C in the alternate arc DA

These two angles are equal (Try to prove this!)

What is 'alternate'?

Notice that the smaller arc of the circle intersects with $\angle BAD$. Alternate arc DA implies we have to consider the other arc (in this case the bigger arc DA) and choose a point C from it.

Use the alternate segment theorem to conclude: $\angle BAD = \angle ACD$

## Hint 3: Draw the circumcircle

This is a major hint!

Draw the circumcircle of ABC.

You should definitely stop reading at this point and try the problem on your own.

## Hint 4: What is the position of E?

Extend AE to meet the circumference of ABC at F.

Can you show that E is the center of this circle?

## Hint 5: E is center

$\angle AED = \angle ACD$ because both of them are subtended by the same arc AB at the circumference.

Hence ABF is isosceles because ($\\angle AED = \angle ACD =\angle BAF$

As BE is perpendicular to the base of the isosceles triangle, it bisects the base implying AE = EF.

We already know AE = CE.

Hence A, C, F are equidistant from E. Hence E is the center.

## Hint 6: Final angle chasing.

Since AF is diameter, $\angle ABF = 90^0$ (angle in the semicircle).

Thus $\angle BAF = 45^0$ implying $\angle BCA = 45^o$

## Reference Book

Challenges and Thrills of Pre-college Mathematics.

# Knowledge Partner  