tangent iff OM=ON

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tangent iff OM=ON

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Source: Iran 3rd round 2011-geometry exam-p2

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#1 h Sep 6, 2011, 10:28 AM • 5 Y

Y by mahanmath, Amir Hossein, Systematicworker, Adventure10, and 1 other user

In triangle $ABC$ , $\omega$ is its circumcircle and $O$ is the center of this circle. Points $M$ and $N$ lie on sides $AB$ and $AC$ respectively. $\omega$ and the circumcircle of triangle $AMN$ intersect each other for the second time in $Q$ . Let $P$ be the intersection point of $MN$ and $BC$ . Prove that $PQ$ is tangent to $\omega$ iff $OM=ON$ .

proposed by Mr.Etesami

This post has been edited 1 time. Last edited by goodar2006, Sep 14, 2011, 5:25 PM

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#2 h Sep 6, 2011, 5:15 PM • 6 Y

Y by Nguyenhuyhoang, Sx763_, Systematicworker, Adventure10, Mango247, and 1 other user

Since $Q$ is the Miquel point of $\triangle ABC \cup MN,$ it follows that $P,Q,M,B$ are concyclic. Let $R \equiv PQ \cap AC$ and assume that $\overline{PQR}$ is tangent to the circumcircle $(O).$ Then $\angle RQA=\angle QBM=\angle QPM$ $\Longrightarrow$ $QA \parallel PN.$ Hence, the cyclic quadrilateral $AQMN$ is an isosceles trapezoid with bases $AQ$ and $MN$ $\Longrightarrow$ perpendicular bisector of $\overline{AQ}$ is also the perpendicular bisector of $\overline{MN}$ $\Longrightarrow$ $OM=ON.$ The converse can be proved with exactly the same arguments.

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sayantanchakraborty

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#3 h Sep 1, 2014, 2:16 PM • 1 Y

Y by Adventure10

This problem is nice.Here's a pure synthetic solution.

1. $PQ$ tangent to $\omega \Rightarrow OM=ON$ .

Since $PQ$ is tangent to $\omega$ we have $\angle{CQP}=\angle{QBC}=\angle{QBP}$ .Also note that $\angle{CQP}=\angle{QAC}=\angle{QAN}=\angle{QMN}=\angle{QMP}$ .Thus $\angle{QMP}=\angle{QBP} \implies QMBP$ is concyclic.So $\angle{NCQ}=\angle{ACQ}=\angle{ABQ}=\angle{QBM}=\angle{QPM}=\angle{QPN} \implies QNCP$ is cyclic.Hence $\angle{CQP}=\angle{CNP}=\angle{ANM}=\angle{QAN}$ .So $AQ \parallel MN \implies AQNM$ is an isosceles trapezoid.Thus $QM=AN$ and $AM=QN$ .

Now note that $\angle{AMQ}=\angle{ANQ} \Rightarrow \angle{BMQ}=\angle{CNQ}$ .Also $\angle{MBQ}=\angle{ABQ}=\angle{ACQ}=\angle{NCQ}$ .Thus $\triangle{BMQ} \sim \triangle{CNQ} \implies \frac{BM}{CN}=\frac{MQ}{NQ} \implies BM \times NQ=CN \times QM \implies BM \times AM=CN \times AN$ .
So since power of $M$ and $N$ are equal wrt $\omega$ we have $OM=ON$ .

2. $OM=ON \Rightarrow PQ$ is tangent to $\omega$ .

As with the above situation it follows that $\triangle{BMQ} \sim \triangle{CNQ} \Rightarrow \frac{BM}{CN}=\frac{MQ}{NQ} \Rightarrow BM \times NQ=CN \times MQ$ .
Also by hypothesis we have $BM \times AM=CN \times AN$ .Combining these yeilds
$AN \times NQ=AM \times MQ \Rightarrow [AMQ]=[ANQ] \Rightarrow [AMZ]=[QNZ] \Rightarrow AZ \times MZ=QZ \times NZ$ .
We also have $\frac{AZ}{MZ}=\frac{QZ}{NZ}$ .Summing these it follows that $AN=QM$ or $AQNM$ is an isosceles trapezoid.

Also note that as with the previous situation we similarly have $QNCP$ cyclic.Thus $\angle{PQC}=\angle{PNC}=\angle{ANM}=\angle{NAQ}=\angle{CAQ}$ or $PQ$ is tangent to $\omega$ as required.

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#4 h Sep 7, 2017, 6:41 AM • 3 Y

Y by ike.chen, Adventure10, Mango247

$[asy] unitsize(100); pen n_purple = rgb(0.7,0.4,1), n_blue = rgb(0,0.6,1), n_green = rgb(0,0.4,0), n_orange = rgb(1,0.4,0.1), n_red = rgb(1,0.2,0.4); pair A, B, C, O, M, N, Q, P; A = dir(110); B = dir(210); C = dir(330); O = origin; M = intersectionpoints(circle(O, 0.7), A--B)[0]; N = intersectionpoints(circle(O, 0.7), A--C)[1]; Q = reflect(O, circumcenter(A, M, N)) * A; P = extension(M, N, B, C); draw(P--Q, gray(0.6)); draw(C--P, gray(0.6)); draw(M--P, gray(0.6)); draw(A--Q, gray(0.6)); draw(A--B--C--cycle); draw(circumcircle(A, M, N), n_red); draw(unitcircle, n_blue); dot(A^^B^^C^^M^^N^^P^^Q); label("$A$", A, dir(130)); label("$B$", B, dir(210)); label("$C$", C, dir(290)); label("$M$", M, dir(180)); label("$N$", N, dir(70)); label("$Q$", Q, dir(340)); label("$P$", P, dir(310)); [/asy]$
By angle chasing, $\overline{PQ}$ is tangent to $\omega$ iff $\overline{AQ} \parallel \overline{MN}$ , so it suffices to show $OM = ON$ iff $\overline{AQ} \parallel \overline{MN}$ .

Lemma. Given two circles $\omega_1$ and $\omega_2$ , the locus of points $P$ such that $\text{Pow}_{\omega_1}(P) - \text{Pow}_{\omega_2}(P)$ is constant is a line perpendicular to the line joining the centers of $\omega_1$ and $\omega_2$ .

Proof. $\text{Pow}_{\omega_1}(P) - \text{Pow}_{\omega_2}(P)$ is linear in $P$ .

Now $\overline{MN}$ is parallel to $\overline{AQ}$ , the radical axis of $\omega$ and $(AMN)$ ,
iff $\text{Pow}_{\omega}(M) - \text{Pow}_{(AMN)}(M) = \text{Pow}_{\omega}(N) - \text{Pow}_{(AMN)}(N)$
iff $\text{Pow}_{\omega}(M) = \text{Pow}_{\omega}(N)$
iff $OM = ON$ .

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#5 h Jan 21, 2019, 3:50 AM • 3 Y

Y by Adventure10, Mango247, Mango247

Notice that $\frac{PB}{PC} = \left(\frac{QB}{QC}\right)^2 = \left(\frac{BM}{CN}\right)^2$ . By Menelaus' theorem on $\triangle ABC$ and $\overline{PMN}$ , we have $\frac{PB}{PC} = \frac{AN}{AM} \cdot \frac{BM}{CN}$ .

Combining the two equations yields $\frac{AN}{BM} = \frac{AM}{CN}$ , which is equivalent to $BM \cdot AM = CN \cdot AN \implies \text{Pow}(M, \odot(O)) = \text{Pow}(N, \odot(O))$ so $OM = ON$ .

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#7 h Nov 15, 2021, 10:09 PM

Y by

Let $\gamma$ denote $(AQMN)$ .

First, assume $PQ$ is tangent to $\omega$ . Since $Q$ is the Miquel Point of complete quadrilateral $BCNM$ , we know $BMQP$ is cyclic. Thus, we have $\measuredangle NMA = \measuredangle PMB = \measuredangle PQB = \measuredangle QAB = \measuredangle QAM$ so $AQ \parallel MN$ . Hence, $AQMN$ is a (cyclic) isosceles trapezoid. Properties of Miquel Points yield $QBM \overset{+}{\sim} QCN$ , so $\frac{AM}{AN} = \frac{QN}{QM} = \frac{CN}{BM}$ which gives $Pow_{\omega}(M) = AM \cdot BM = AN \cdot CN = Pow_{\omega}(N)$ clearly implying $OM = ON$ .

Now, assume $OM = ON$ . Reversing our logic clearly gives $\frac{AM}{AN} = \frac{CN}{BM} = \frac{QN}{QM}.$ Since $M$ is in between $A$ and $B$ , and $N$ is in between $A$ and $C$ , we know that neither of these points can lie outside $(ABC)$ . Because one of the arcs of $\gamma$ bounded by $A$ and $Q$ must travel outside $(ABC)$ , it follows that $A$ and $Q$ are adjacent vertices of quadrilateral $AQMN$ . Since there exists exactly one point $Q$ on $\gamma$ which is on the same side of $MN$ as $A$ such that $\frac{AM}{AN} = \frac{QN}{QM}$ it follows that $Q$ is the unique point on $\gamma$ such that $AQ \parallel MN$ . Now, angle chasing clearly implies the desired tangency, so we're done. $\blacksquare$

Remarks: My solution is unnecessarily long and not generalizable because of configuration issues. Recognizing the perpendicular bisector argument is much cleaner.

Overall, this was a pretty easy problem.

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#8 h Jul 31, 2023, 3:36 PM

Y by

Very nice problem. The condition is equivalent to $\frac{BM}{CN} = \frac{AN}{AM}$ by Power of a Point, which rewrites as $\triangle QMN \sim \triangle ANM$ by considering $Q$ as the Miquel point. On the other hand, this implies $AQNM$ is an isosceles trapezoid. As $QNCP$ is also cyclic, we have $\measuredangle PQC = \measuredangle QNM = \measuredangle QAC,$ which is equivalent to the tangency.

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