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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>:
 
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>.
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<math>Kris17</math>
 
<math>Kris17</math>
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== Solution 2 (By Burapat1729) ==
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Let <math>A_H \bot BC</math> to show <math>A_H,K,A</math> are colinear.
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Easy to see that, <math>Q,E,A_H,K</math> and <math>A_H,K,F,P</math> concylic.
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Then, <math>KA_H</math> be radical axis of those two circle.
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So, midpoint of <math>EK</math> and <math>FK</math> is center of those two circle. Let be <math>m_1,m_2</math>
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So by normal similarity, <math>m_1m_2\parallel EF \parallel BC</math>.
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By radical axis lemma, <math>KA_H \bot m_1m_2 \longleftrightarrow KA_H \bot BC</math>. Therefore <math>A_H,K,A</math> are colinear. Q.E.D

Latest revision as of 10:18, 11 October 2024

Problem 3

A half-circle of diameter $EF$ is placed on the side $BC$ of a triangle $ABC$ and it is tangent to the sides $AB$ and $AC$ in the points $Q$ and $P$ respectively. Prove that the intersection point $K$ between the lines $EP$ and $FQ$ lies on the altitude from $A$ of the triangle $ABC$.


Solution

We begin by showing that $A$ is the circumcenter of $\triangle KPQ$:

Consider the configuration where $\angle ACB$ is obtuse and diameter $EF$ lies on extended line $BC$.

Let $O$ be the midpoint of diameter $EF$. Let $KA$ meet $BC$ at $G$.

Let us define $\angle KPA = \alpha$ and $\angle KQA = \beta$

By applying Tangent Chord Angle theorem, we get: $\angle POE = 2\alpha$ and $\angle QOF = 2\beta$

Now, $\angle POQ = 180 - 2(\alpha + \beta)$, and since $PAQO$ is a cyclic quadrilateral, we have $\angle PAQ = 2(\alpha + \beta)$

Now $\angle POF = 180 - 2\alpha$, so $\angle PEF = 90 - \alpha$

Similarly, we have $\angle QOE = 180 - 2\beta$, so $\angle QFE = 90 - \beta$

From $\triangle EKF, \angle EKF = 180 - (\angle PEF + \angle QFE)$

$= 180 - (90 - \alpha + 90 - \beta) = (\alpha + \beta)$

Thus, we have $\angle PKQ = \angle EKF = \angle PAQ/2$

Also, $AP = AQ$ (Since $AP$ and $AQ$ are tangents to the same circle)

From the above 2 results, it readily follows that $A$ is the circumcenter of $\triangle KPQ$.

Thus, we have $AK = AP$, and so $\angle AKP = \angle KPA = \alpha$

So in $\triangle EKG, \angle KGE = 180 - (\angle AKP + \angle PEF)$

$= 180 - (\alpha + 90 - \alpha) = 90^{\circ}$

So $KA$ is perpendicular to $BC$, hence $K$ lies on the altitude from $A$ of the triangle $ABC$.


$Kris17$

Solution 2 (By Burapat1729)

Let $A_H \bot BC$ to show $A_H,K,A$ are colinear.

Easy to see that, $Q,E,A_H,K$ and $A_H,K,F,P$ concylic.

Then, $KA_H$ be radical axis of those two circle.

So, midpoint of $EK$ and $FK$ is center of those two circle. Let be $m_1,m_2$

So by normal similarity, $m_1m_2\parallel EF \parallel BC$.

By radical axis lemma, $KA_H \bot m_1m_2 \longleftrightarrow KA_H \bot BC$. Therefore $A_H,K,A$ are colinear. Q.E.D

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