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Difference between revisions of "1990 USAMO Problems/Problem 5"

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An acute-angled triangle <math>ABC</math> is given in the plane. The circle with diameter <math>\, AB \,</math> intersects altitude <math>\, CC' \,</math> and its extension at points <math>\, M \,</math> and <math>\, N \,</math>, and the circle with diameter <math>\, AC \,</math> intersects altitude <math>\, BB' \,</math> and its extensions at <math>\, P \,</math> and <math>\, Q \,</math>. Prove that the points <math>\, M, N, P, Q \,</math> lie on a common circle.
 
An acute-angled triangle <math>ABC</math> is given in the plane. The circle with diameter <math>\, AB \,</math> intersects altitude <math>\, CC' \,</math> and its extension at points <math>\, M \,</math> and <math>\, N \,</math>, and the circle with diameter <math>\, AC \,</math> intersects altitude <math>\, BB' \,</math> and its extensions at <math>\, P \,</math> and <math>\, Q \,</math>. Prove that the points <math>\, M, N, P, Q \,</math> lie on a common circle.
  
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== Solution ==
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== Solution 1==
 
Let <math>A'</math> be the intersection of the two circles (other than <math>A</math>). <math>AA'</math> is perpendicular to both <math>BA'</math>, <math>CA'</math> implying <math>B</math>, <math>C</math>, <math>A'</math> are collinear. Since <math>A'</math> is the foot of the altitude from <math>A</math>: <math>A</math>, <math>H</math>, <math>A'</math> are concurrent, where <math>H</math> is the orthocentre.
 
Let <math>A'</math> be the intersection of the two circles (other than <math>A</math>). <math>AA'</math> is perpendicular to both <math>BA'</math>, <math>CA'</math> implying <math>B</math>, <math>C</math>, <math>A'</math> are collinear. Since <math>A'</math> is the foot of the altitude from <math>A</math>: <math>A</math>, <math>H</math>, <math>A'</math> are concurrent, where <math>H</math> is the orthocentre.
  

Revision as of 17:33, 19 July 2017

Problem

An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $\, AB \,$ intersects altitude $\, CC' \,$ and its extension at points $\, M \,$ and $\, N \,$, and the circle with diameter $\, AC \,$ intersects altitude $\, BB' \,$ and its extensions at $\, P \,$ and $\, Q \,$. Prove that the points $\, M, N, P, Q \,$ lie on a common circle.

Solution 1

Let $A'$ be the intersection of the two circles (other than $A$). $AA'$ is perpendicular to both $BA'$, $CA'$ implying $B$, $C$, $A'$ are collinear. Since $A'$ is the foot of the altitude from $A$: $A$, $H$, $A'$ are concurrent, where $H$ is the orthocentre.

Now, $H$ is also the intersection of $BB'$, $CC'$ which means that $AA'$, $MN$, $PQ$ are concurrent. Since $A$, $M$, $N$, $A'$ and $A$, $P$, $Q$, $A'$ are cyclic, $M$, $N$, $P$, $Q$ are cyclic by the radical axis theorem.

Solution 2

Define $A'$ as the foot of the altitude from $A$ to $BC$. Then, $AA' \cap BB' \cap CC'$ is the orthocenter. We will denote this point as $H$ Since $\angle AA'C$ and $\angle AA'B$ are both $90^{\Circle}$, $A'$ lies on the circles with diameters $AC$ and $AB$.

Now we use the Power of a Point theorem with respect to point $H$. From the circle with diameter $AB$ we get $AH \cdot A'H = MH \cdot NH$. From the circle with diameter $AC$ we get $AH \cdot A'H = PH \cdot QH$. Thus, we conclude that $PH \cdot QH = MH \cdot NH$, which implies that $P$, $Q$, $M$, and $N$ all lie on a circle.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1990 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Last Question
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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