Pascal's Theorem

The intersection of chords formed by six points are collinear.

Pascal's theorem is a very useful theorem in Olympiad geometry to prove the collinearity of three intersections among six points on a circle.

The theorem states as follows:

Pascal's Theorem

Given 6 points (which can be coincident) on the circumference of a circle labelled $A, C, E, B, F,$ and $D$ in that order around the circle, the intersections of $AB$ and $DE$, $AF$ and $CD$, and $BC$ and $EF$ are collinear.

There are many different ways to prove this theorem, but an easy way is to use Menelaus' theorem.

Let $G$ be the intersection of $\overline{CD}$ and $\overline{FA},$ let $H$ be the intersection of $\overline{AB}$ and $\overline{DE},$ and let $I$ be the intersection of $\overline{BC}$ and $\overline{EF}.$ We will prove these three points are collinear.

Let $U$ be the intersection of $\overline{CD}$ and $\overline{EF},$ let $V$ be the intersection of $\overline{AB}$ and $\overline{EF},$ and let $W$ be the intersection of $\overline{AB}$ and $\overline{CD}.$ By Menelaus in $\triangle UVW$ and line $HDE$, we have

$\dfrac {VH}{WH} \cdot \dfrac {WD}{UD} \cdot \dfrac {UE}{VE} = 1.$

By Menelaus in $\triangle UVW$ and line $AGF$, we have

$\dfrac {VA}{WA} \cdot \dfrac {WG}{UG} \cdot \dfrac {UF}{VF} = 1.$

By Menelaus in $\triangle UVW$ and line $BCI$, we have

$\dfrac {VB}{WB} \cdot \dfrac {WC}{UC} \cdot \dfrac {UI}{VI} = 1.$

Multiplying these out, we get

$\dfrac {VH}{WH} \cdot \dfrac {WD}{UD} \cdot \dfrac {UE}{VE} \cdot \dfrac {VA}{WA} \cdot \dfrac {WG}{UG} \cdot \dfrac {UF}{VF} \cdot \dfrac {VB}{WB} \cdot \dfrac {WC}{UC} \cdot \dfrac {UI}{VI} = 1.$

Upon rearranging, we get

$\dfrac {WD}{UD} \cdot \dfrac {UE}{VE} \cdot \dfrac {VA}{WA} \cdot \dfrac {UF}{VF} \cdot \dfrac {VB}{WB} \cdot \dfrac {WC}{UC} \cdot \dfrac {VH}{WH} \cdot \dfrac {WG}{UG} \cdot \dfrac {UI}{VI} = 1.$

Notice that by power of a point, we have

$\begin{aligned} \dfrac {WD}{UD} \cdot \dfrac {UE}{VE} \cdot \dfrac {VA}{WA} \cdot \dfrac {UF}{VF} \cdot \dfrac {VB}{WB} \cdot \dfrac {WC}{UC} &= \dfrac {WD \times WC}{WA \times WB} \cdot \dfrac {VA \times VB}{VE \times VF} \cdot \dfrac {UE \times UF}{UC \times UD}\\\\ &= 1. \end{aligned}$

Thus, the product above simplifies as

$\dfrac {VH}{WH} \cdot \dfrac {WG}{UG} \cdot \dfrac {UI}{VI} = 1.$

So, by Menelaus, $G, H,$ and $I$ are collinear. $_\square$

Consider a circle internally tangent to the circumcircle of triangle $ABC$ and also tangent to sides $\overline{AB}$ and $\overline{AC}$ at $P$ and $Q,$ respectively. Show that $\overline{PQ}$ passes through the incenter $I$ of $\triangle ABC.$

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At a glance, we can see that we want $P, I, Q$ to be the points of collinearity, but this may not seem obvious at first, especially since there are only 3 points on the circumcircle. However, if we break this question down into small, manageable chunks, we can break through with solving this.

Let $T$ be the point of tangency between the circumcircle and the smaller circle (this circle is called the mixtilinear circle). We extend $TP$ to meet the circumcircle again at $X$. We claim that $BX = AX$, and we prove it as follows:

Let $TB$ intersect the mixtilinear circle again at $R$, and let $O$ be the centre of the mixtilinear circle. By alternate segment theorem, $\angle RPT$ is equal to the angle made by $BT$ and the tangent, which is equal to $\angle BXT$, so $RP \parallel BX$. We will call $\angle OTP = \alpha$ and $\angle RTP = \beta$. The following angle chase is not too hard and is left as an exercise for the reader. Angle chasing results in $\angle BXT = 90^{\circ} - \alpha - \beta$, $\angle XBT = 90^{\circ} + \alpha$ and $\angle AXT = \angle ABT = 90^{\circ} + \alpha - \beta$, from which we get $\angle AXB = 180^{\circ} - 2\beta$ and $\angle XBA = \beta$, so $\angle XAB = \beta$. Thus, $\Delta XAB$ is isosceles, so $BX=AX$.

Thus, our claim is proven. From this statement, we can invoke the following property:

If $X$ is a point on the circumcircle of $ABC$ on the minor arc $BC,$ $BX=XC$ if and only if $AX$ is the angle bisector of $\angle BAC$.

From this property, we get $CX$ is the angle bisector of $\angle ACB$. Similarly, if $TQ$ is extended to $Y$, $AY = CY$, so $BY$ is the angle bisector of $\angle ABC$. Now we can finally apply Pascal's theorem. By Pascal's theorem in $XCABYT$ (following the line segments in that order and cycling back), we get $P$, $I$ and $Q$ as the collinear intersection points, so we are done. $_\square$

Prove the fundamental theorems of poles and polars:

Given a cyclic quadrilateral $ABCD$, if the intersection of $\overline{AC}$ and $\overline{BD}$ is $P$, the intersection of $\overline{AB}$ and $\overline{CD}$ is $Q$, and the intersection of $\overline{AD}$ and $\overline{BC}$ is $R$, prove the polar of $P$ passes through $Q$ and $R$.

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Firstly, here are some definitions. Let $O$ be the center of a circle with radius $R$, and $P$ an arbitrary point, which we will call the pole. Let $P'$ be the point on $OP$ (possibly extended) such that $OP \times OP' = R^2$. The line through $P'$ perpendicular to $OP$ is known as the polar of $P$. A unique property (left to be proven by the reader) of poles and polars is that if $P$ is inside the circle, then $P'$ is the intersection of the tangents from the endpoints of the chord with midpoint $P$.

The way to solve this problem is with Pascal's. Note that Pascal's can be applied even if two or more points are coincident. Let us consider Pascal's in hexagon $ACCBDD$. Then, $AC \cap BD = P$, $CC \cap DD$ (the line through coincident points would be a tangent to the circle) and $CB \cap DA = R$ are collinear. Similarly, if we use Pascal's in $CAADBB$, we have $CA \cap DB = P$, $AA \cap BB = H$ and $AD \cap BC = R$. Thus, $P$, $R$, $AA \cap BB$ and $CC \cap DD$ are collinear. A similar argument shows $P$, $Q$, $BB \cap CC = I$ and $DD \cap AA$ are collinear. From this, using the fact that $AA$ is a tangent, etc., we get that $QR$ is parallel to $HI$, from which we can gather $QR$ is a pole of $P$. $_\square$