Theorems in Geometry

Theorems May 26, 2024

Around 300 BC, the Greek mathematician, Euclid, takes on a massive project to summarize all mathematics known at the time, essentially creating a series of books containing everything everyone knows about math. Since then, Euclid's treatise "Elements" has been published in more editions than any other book except the Bible.

Book 1 contains 5 postulates that derive all modern theorems in flat geometry. It defines many figures living inside the 2D plane along with relationships among themselves, some of which you might already be familiar with:

There are two common measures in geometry: angle and distance. The infamous fifth postulate especially talks about the angle between two lines, and it's the reason why Euclidean geometry is flat. Proving a geometric statement involving angles can be a pain in the ass since lines derived from one statement can be interpreted in multiple configurations.

To address this issue, mathematicians come up with directed angle. The directed angle \(\measuredangle ABC\) is defined as the angle of counter-clockwise rotation in modulo \(180^{\circ}\) transforming line \(AB\) to line \(BC\). Here are some trivial facts rewritten in directed angles:

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【Angle Addition Postulate】For any points \(O\), \(A\), \(P\), \(B\), we have \(\measuredangle AOP + \measuredangle POB = \measuredangle AOB\).

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【Triangles Sum to \(180^{\circ}\)】For any points \(A\), \(B\), \(C\), we have \(\measuredangle ABC + \measuredangle BCA + \measuredangle CAB = 0\).

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【Collinearity Criteria】Let \(X\) be any point. Points \(A\), \(B\), \(C\) are collinear if and only if \(\measuredangle XBC = \measuredangle XBA\).

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【Directed Cyclic Quadrilateral】Let \(A\), \(B\), \(X\), \(Y\) be four points, no three are collinear. They are concyclic if and only if \(\measuredangle XAY = \measuredangle XBY\).

Spiral Similarity

Two figures are called similar if they have the same shape. This means one can be transformed from the other by dilation, translation, rotation, or reflection. If the transformation has no dilation, the two are called congruent and will have the same size.

Now, spiral similarity is a transformation \(\textbf{T}\) composed of only a rotation and a dilation centered at the same point \(O\) with coordinate \(z_0\). By definition, for any point \(z\) in the complex plane, \(\textbf{T}\) satisfies \[\textbf{T}(z) = z_0 + \alpha (z − z_0),\] where \(|\alpha|\) is the dilation factor, and \(\arg(\alpha)\) is the angle of rotation.

Consider four distinct points \(A\), \(B\), \(C\), \(D\) in the plane with complex coordinates \(a\), \(b\), \(c\), \(d\), respectively. Let's look for transformations that sends \(A\) to \(C\), and \(B\) to \(D\).

If \(a - c = b - d\), then \(AC \parallel BD\) and the translation \(a \mapsto c\) would be one such transformation.
If \(a - c \neq b - d\), then the unique solution \(\alpha = \frac{c - d}{a - b}\) and \(z_0 = \frac{ad - bc}{a - b - c + d}\) satisfy the spiral equation

\begin{aligned}
\textbf{T}(a) &= z_0 + \alpha (a − z_0) = c \\
\textbf{T}(b) &= z_0 + \alpha (b − z_0) = d,
\end{aligned}

and hence the spiral similarity \(\textbf{T}\) would be another transformation we found. Note that the existence of \(\textbf{T}\) implies \(\triangle OAB \sim \triangle OCD\), and vice versa. We conclude:

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Given four arbitrary distinct points \(A\), \(B\), \(C\), \(D\), there exists either a translation or a unique spiral similarity that sends \(A\) to \(C\), and \(B\) to \(D\).

Now, geometrically speaking, how do you find the center \(z_0\) of the spiral similarity? We'll explain this in the theorem below.

Let \(A\), \(B\), \(C\), \(D\) be four distinct points in the plane such that \(AC \nparallel BD\). Let lines \(AC\) and \(BD\) meet at \(X\).

If the circumcircles of \(\triangle ABX\) and \(\triangle CDX\) meet again at \(O\), then \(O\) is the center of the spiral similarity that carries \(A\) to \(C\) and \(B\) to \(D\).
Conversely, if a point \(O\) is the center of the spiral similarity that carries \(A\) to \(C\) and \(B\) to \(D\), then \(ABXO\) and \(CDXO\) are cyclic quadrilaterals.

Proof

To avoid configuration issues, we will use directed angles. You can play with the image below to see such different configurations.

Assuming the first statement is proved, the uniqueness of spiral similarity will imply the second statement. Hence we only need to prove the former. So, assume \(ABXO\) and \(CDXO\) are cyclic quadrilaterals. We have \begin{aligned} \measuredangle OAB &= \measuredangle OXB = \measuredangle OXD = \measuredangle OCD \\ \measuredangle OBA &= \measuredangle OXA = \measuredangle OXC = \measuredangle ODC, \end{aligned} and hence \(\triangle OAB \sim \triangle OCD\). In other words, \(O\) is the center of the spiral similarity that carries \(A\) to \(C\) and \(B\) to \(D\).

\(\blacksquare\)

Power of a Point

Angles are cool, but let's talk about distance. You can easily calculate the distance between two points by measuring the length of a straight segment connecting both. In 1826, a Swiss mathematician, Jakob Steiner, was once tasked with a more interesting thought experiment: how do you calculate the "distance" between a point and a circle? Which segment do you even measure?

Consider a point \(P\) and a circle \(\omega\). Let's call their distance \(\Pi(P, \omega)\). By definition, \(\Pi\) should be:

constant if \(P\) and \(\omega\) are not moving (just like the constant distance between two static points), and
constant regardless the choice of the segment connecting \(P\) and \(\omega\).

By letting \(O\) and \(r\) be the center and radius of \(\omega\), respectively, Steiner called \(\Pi\) the power of \(P\) with respect to \(\omega\) and define it as \[\Pi(P, \omega) := |\vec{PO}|^2 - r^2.\]

It's easy to see that the first requirement above holds. The second one will be proved in the following theorem.

Let \(P\) be a point and \(\omega\) be a circle with radius \(r\) centered at \(O\). A line through \(P\) intersects \(\omega\) at \(A\) and \(B\) (if \(A = B\), then the line touches \(\omega\)). Then we have \[\Pi(P, \omega) := |\vec{PO}|^2 - r^2 = \vec{PA} \cdot \vec{PB}.\]

Proof

If \(A = B\), the result follows directly from the Pythagorean theorem. If \(A \neq B\), WLOG, \(A\) is closer to \(P\) than \(B\) is. Let the line through \(P\) and \(O\) intersects \(\omega\) at \(C\) and \(D\), where \(C\) is closer to \(P\) than \(D\) is. Since ABCD is cyclic, we get \[\measuredangle PAC = \measuredangle BAC = \measuredangle BDC = \measuredangle BDP.\]

Since we also know \(\measuredangle CPA = \measuredangle DPB\), then \(\triangle PAC \sim \triangle PDB\). Hence, \[\vec{PA} \cdot \vec{PB} = \vec{PC} \cdot \vec{PD} = (\vec{PO} - \vec{OD})(\vec{PO} + \vec{OD}) = |\vec{PO}|^2 - |\vec{OD}|^2 = |\vec{PO}|^2 - r^2.\]

\(\blacksquare\)

If \(P\) is inside \(\omega\), then \(\Pi\) is negative, so \(\Pi\) is not really a distance. However, in many Olympiad problems, we care more about the constant value of \(\Pi\) rather than the value itself. Moreover, since \(\Pi\) is the dot product of two vectors with the same or opposite direction, we usually ignore the direction and eliminate the negative sign. In other words, in the proof above we're more interested in the relation \(PA \cdot PB = PC \cdot PD\) than the value of the multiplication itself.

Every "theorem" post such as this one will be updated progressively as we solve more and more problems. Stay tuned!