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

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Now we use the Power of a Point theorem with respect to point <math>H</math>. From the circle with diameter <math>AB</math> we get <math>AH \cdot A'H = MH \cdot NH</math>. From the circle with diameter <math>AC</math> we get <math>AH \cdot A'H = PH \cdot QH</math>. Thus, we conclude that <math>PH \cdot QH = MH \cdot NH</math>, which implies that <math>P</math>, <math>Q</math>, <math>M</math>, and <math>N</math> all lie on a circle.
 
Now we use the Power of a Point theorem with respect to point <math>H</math>. From the circle with diameter <math>AB</math> we get <math>AH \cdot A'H = MH \cdot NH</math>. From the circle with diameter <math>AC</math> we get <math>AH \cdot A'H = PH \cdot QH</math>. Thus, we conclude that <math>PH \cdot QH = MH \cdot NH</math>, which implies that <math>P</math>, <math>Q</math>, <math>M</math>, and <math>N</math> all lie on a circle.
  
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{{alternate solutions}}
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==Solution 3 (Radical Lemma)==
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Let <math>\omega_1</math> be the circumcircle with diameter <math>AB</math> and <math>\omega_2</math> be the circumcircle with diameter <math>AC</math>.  We claim that the second intersection of <math>\omega_1</math> and <math>\omega_2</math> other than <math>A</math> is <math>A'</math>, where <math>A'</math> is the feet of the perpendicular from <math>A</math> to segment <math>BC</math>.  Note that <cmath>\angle AA'B=90^{\circ}=\angle AB'B</cmath> so <math>A'</math> lies on <math>\omega_1.</math>  Similarly, <math>A'</math> lies on <math>\omega_2</math>.  Hence, <math>AA'</math> is the radical axis of <math>\omega_1</math> and <math>\omega_2</math>.  By the Radical Lemma, it suffices to prove that the intersection of lines <math>MN</math> and <math>PQ</math> lie on <math>AA'</math>.  But, <math>MN</math> is the same line as <math>CC'</math> and <math>PQ</math> is the same line as <math>BB'</math>.  Since <math>AA', BB'</math>, and <math>CC'</math> intersect at the orthocenter <math>H</math>, <math>H</math> lies on the radical axis <math>AA'</math> and we are done. <math>\blacksquare</math>
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==Solution 4==
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We know that <math>BCB'C'</math> is a cyclic quadrilateral. Hence,
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<math>HB \cdot HB' = HC \cdot HC' </math>
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<math>\implies Pow_{\omega_{1}} = Pow_{\omega_{2}} </math>
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<math>\implies HM \cdot HN = HP \cdot HQ </math>
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<math>\implies MPNQ </math> is cyclic <math>\raggedright\blacksquare </math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 00:42, 27 September 2024

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 $90^{\circ}$, $A'$ 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.

Solution 3 (Radical Lemma)

Let $\omega_1$ be the circumcircle with diameter $AB$ and $\omega_2$ be the circumcircle with diameter $AC$. We claim that the second intersection of $\omega_1$ and $\omega_2$ other than $A$ is $A'$, where $A'$ is the feet of the perpendicular from $A$ to segment $BC$. Note that \[\angle AA'B=90^{\circ}=\angle AB'B\] so $A'$ lies on $\omega_1.$ Similarly, $A'$ lies on $\omega_2$. Hence, $AA'$ is the radical axis of $\omega_1$ and $\omega_2$. By the Radical Lemma, it suffices to prove that the intersection of lines $MN$ and $PQ$ lie on $AA'$. But, $MN$ is the same line as $CC'$ and $PQ$ is the same line as $BB'$. Since $AA', BB'$, and $CC'$ intersect at the orthocenter $H$, $H$ lies on the radical axis $AA'$ and we are done. $\blacksquare$

Solution 4

We know that $BCB'C'$ is a cyclic quadrilateral. Hence,

$HB \cdot HB' = HC \cdot HC'$

$\implies Pow_{\omega_{1}} = Pow_{\omega_{2}}$

$\implies HM \cdot HN = HP \cdot HQ$

$\implies MPNQ$ is cyclic $\raggedright\blacksquare$.

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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