Find the measure of $∠OTS$ in a circle

Question

I need help with the following Geometry problem. In a circle with diameter $AB$ and center $O$, chords $PQ$ and $QC$ are drawn that intersect $AB$ at $M$ and $N$, respectively. If the extensions of $PC$ and $QO$ intersect line $AB$ and the circle at points $S$ and $T$, in that order, and $MO=ON$, find the measure of $∠OTS$. Diagram

For context, this problem was proposed by my geometry teacher. These are problems we use to practice before taking the admission exam held here in Peru to enter the National University of Engineering (UNI).

I tried to use the given $MO = ON$ to look for an application of Thales' Theorem, but I couldn't find any construction that would help significantly. Another idea I had was to draw segment $PO$ and, since it's a median of segment $NM$, locate the centroid, and since $PO$ is also equal to the radius of the circle, take advantage of the proportions in $AO$.

I used Geogebra to see what the angle value would be regardless of the positions of $P$, $Q$, and $C$, and it showed that the angle is $90°$. So I tried to look for the relationship between segments $ST$, $TO$, and $SO$ to prove it, but I didn't get anywhere.

It would help me a lot if you could at least share your ideas on how you would solve the problem or what steps I could follow to solve it. I hope my request is well received and that you find the problem interesting. Thank you very much.

P.S. I'm sorry if this problem is rather simple. I know it doesn't compare to the many much more difficult problems on this site, but I would truly appreciate the help.

Answer 1

When $OM=ON$ it means we can construct circle d on MN it intersects segment QT at points $A_1$ and $B_1$. Now we construct trapezoid $A_1B_1N_1K$ in the circle c(Circumcircle of triangle ABC) we have $OR\bot N_1K$, this means R is the midpoint of $N_1K$, so $OR$ intersects QS at I. Since $N_1K||QT$ it results in:

$\overset {\Large\frown}{TN_1}=\overset {\Large\frown}{KQ}\Rightarrow \angle N_1QT=\angle KTQ$

This means triangle ITQ is isosceles and also IO is perpendicular to TQ and we can construct circle e on T and Q with center I. This circle passes through S(You have to say the reason), so QS is the diameter of this circle and the opposite angle to it is ninety degrees.that is:

$\angle QTS=90^o$

Update: The reason for passing the circle e through S is:

In right angle triangle SZB we have:

$ZA^2=SA\times AB$

If $\angle STQ=90^o$ then in right angle triangle SZB we have:

$TN_1^2=SN_1\times N_1Q$

we also have the power of S to circle c:

$SA\times SB=SN_1\times SQ$

So we may conclude that triangle STQ is the result of rotation of triangle SZB over S. So S is on the circle e.

Answer 2

To have a clear definition of the used points, and of the used notations, here is an equivalent restatement:

Proposition: On a circle with center $O$ and radius $R$ consider two diameters, $AB$ and $QT$ building an angle $\alpha$ as in the picture. The tangent in $T$ to the circle intersects $AB$ in a point $\Sigma$. On the line $AB$ consider two different points $M,N$ with the same distance $x$ to $O$. The lines $QM$, $QN$ intersect for a second time the circle in points $P$, respectively $C$, $P,C\ne Q$.

Then the points $\Sigma,C,P$ are collinear.

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Proof: (All segments are not signed in the calculations below, the sign is adapted to correspond to the lengths of the segments from this one picture. The reader who knows about this sensible point can please adjust.)

Observe that the points $M,N$ have the same power $\Pi^2:=R^2-x^2$ w.r.t. the circle, for example $MQ\cdot MP=MA\cdot MB=(R+x)(R-x)=(R^2-x^2)$. So we can express $MP,NC$ in terms of $MQ,NQ$ (using $\Pi$), and the squares of these last segments can be expressed by the generalized form of Pythagoras' theorem in terms of $R,x,\alpha$. By construction $R:\Sigma O=TO:\Sigma O=\cos \alpha$. So all needed ingredients have formulas in terms of $R,x,\cos\alpha$, and for the wanted collinearity we apply the reciprocal of Menelaus after the following calculus: $$ \begin{aligned} \frac{\Sigma N}{\Sigma M}\cdot \frac{PM}{PQ}\cdot \frac{CQ}{CN} &= \frac {\displaystyle \frac R{\cos\alpha}-x} {\displaystyle \frac R{\cos\alpha}+x} \cdot \frac {\displaystyle \frac{\Pi^2}{MQ}} {\displaystyle \frac{\Pi^2}{MQ}+MQ} \cdot \frac {\displaystyle \frac{\Pi^2}{NQ}+NQ} {\displaystyle \frac{\Pi^2}{NQ}} \\ &= \frac {\displaystyle R-x\cos\alpha} {\displaystyle R+x\cos\alpha} \cdot \frac {\displaystyle \color{gray}{\Pi^2}} {\displaystyle \Pi^2+MQ^2} \cdot \frac {\displaystyle \Pi^2+NQ^2} {\displaystyle \color{gray}{\Pi^2}} \\ &= \frac {\displaystyle R-x\cos\alpha} {\displaystyle R+x\cos\alpha} \cdot \frac {\displaystyle (R^2-x^2)+(R^2+x^2+2 Rx\cos\alpha)} {\displaystyle (R^2-x^2)+(R^2+x^2-2 Rx\cos\alpha)} \\ &=1\ . \end{aligned} $$ $\square$

Answer 3

So my teacher showed us his method for solving this problem. First, segment $CD$ is drawn, intersecting $TO$ at point $E$ and parallel to $SB$. We ∠$PSM$=β and ∠$SPM$=α. Due to the parallel lines, ∠$PCD$=β. Since $CD$||$SB$, by proportionality, $NO$=$MO$ ⇒ $CE$ = $ED$. We then draw segment $OF$⊥$PC$. Given that $CE$ = $ED$, we draw segment $FE$, which is the midsegment of triangle $CPD$, therefore $FE$||$PD$. Angle $CFE$ = α due to parallel lines. Arc $CAQ$ = 2α, so ∠$CTQ$ = α. We observe that quadrilateral $CTFE$ is cyclic, therefore angle ETF = β. We also note that quadrilateral $STFO$ is cyclic, therefore ∠$STQ$=$90$°.

Answer 4

There are many wonderful solutions before,but I'd like to show mine.

Let $QR$ parallel to $AB$ meet the circle again at $R$.

Then $Q(M,N;O,R)$ is a harmonic bundle, so $RCTP$ is a harmonic quadrilateral

As $R$ is the reflection point of $T$ wrt $AB$, so the tangents from $R,T$ meet at $S'\in AB$

As $RTCP$ is a harmonic quadrilateral, we get $S'\in CP$, so $S' \equiv S$, which means $\angle OTS=90^\circ$