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

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 $\raggedright\blacksquare$ .

@@ Line 22: / Line 22: @@
 We know that <math>BCB'C'</math> is a cyclic quadrilateral. Hence,
-<cmath>HB \cdot HB' = HC \cdot HC' </cmath>
+<math>HB \cdot HB' = HC \cdot HC' </math>
-<cmath>\implies Pow_{\omega_{1}} = Pow_{\omega_{2}} </cmath>
+<math>\implies Pow_{\omega_{1}} = Pow_{\omega_{2}} </math>
-<cmath>\implies HM \cdot HN = HP \cdot HQ </cmath>
+<math>\implies HM \cdot HN = HP \cdot HQ </math>
-<cmath>\implies MPNQ is cyclic \blacksquare </cmath>.
+<math>\implies MPNQ </math> is cyclic <math>\raggedright\blacksquare </math>.
 == 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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