# Difference between revisions of "2000 JBMO Problems/Problem 3"

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== Solution == | == Solution == | ||

+ | We begin by showing that <math>A</math> is the circumcenter of <math>\triangle KPQ</math>: | ||

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+ | Consider the configuration where <math>\angle ACB</math> is obtuse and diameter <math>EF</math> lies on extended line <math>BC</math>. | ||

Let <math>O</math> be the midpoint of diameter <math>EF</math>. | Let <math>O</math> be the midpoint of diameter <math>EF</math>. | ||

Let <math>KA</math> meet <math>BC</math> at <math>G</math>. | Let <math>KA</math> meet <math>BC</math> at <math>G</math>. | ||

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Let us define <math>\angle KPA = \alpha</math> and <math>\angle KQA = \beta</math> | Let us define <math>\angle KPA = \alpha</math> and <math>\angle KQA = \beta</math> |

## Latest revision as of 00:35, 5 December 2018

## Problem 3

A half-circle of diameter is placed on the side of a triangle and it is tangent to the sides and in the points and respectively. Prove that the intersection point between the lines and lies on the altitude from of the triangle .

## Solution

We begin by showing that is the circumcenter of :

Consider the configuration where is obtuse and diameter lies on extended line .

Let be the midpoint of diameter . Let meet at .

Let us define and

By applying Tangent Chord Angle theorem, we get: and

Now, , and since is a cyclic quadrilateral, we have

Now , so

Similarly, we have , so

From

Thus, we have

Also, (Since and are tangents to the same circle)

From the above 2 results, it readily follows that is the circumcenter of .

Thus, we have , and so

So in

So is perpendicular to , hence lies on the altitude from of the triangle .