Pascal's Theorem

Pascal's theorem is a useful theorem in Olympiad geometry to prove the collinearity of three points. The theorem states as follows:

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

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

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 \(P\). \(_\square\)