Subfield: Geometry


Question: Let $T_{1}, T_{2}, T_{3}, T_{4}$ be pairwise distinct collinear points such that $T_{2}$ lies between $T_{1}$ and $T_{3}$, and $T_{3}$ lies between $T_{2}$ and $T_{4}$. Let $\omega_{1}$ be a circle through $T_{1}$ and $T_{4}$; let $\omega_{2}$ be the circle through $T_{2}$ and internally tangent to $\omega_{1}$ at $T_{1}$; let $\omega_{3}$ be the circle through $T_{3}$ and externally tangent to $\omega_{2}$ at $T_{2}$; and let $\omega_{4}$ be the circle through $T_{4}$ and externally tangent to $\omega_{3}$ at $T_{3}$. A line crosses $\omega_{1}$ at $P$ and $W, \omega_{2}$ at $Q$ and $R, \omega_{3}$ at $S$ and $T$, and $\omega_{4}$ at $U$ and $V$, the order of these points along the line being $P, Q, R, S, T, U, V, W$. Prove that $P Q+T U=R S+V W$.

Solution:
Let $O_{i}$ be the centre of $\omega_{i}, i=1,2,3,4$. Notice that the isosceles triangles $O_{i} T_{i} T_{i-1}$ are similar (indices are reduced modulo 4), to infer that $\omega_{4}$ is internally tangent to $\omega_{1}$ at $T_{4}$, and $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ is a (possibly degenerate) parallelogram.



Let $F_{i}$ be the foot of the perpendicular from $O_{i}$ to $P W$. The $F_{i}$ clearly bisect the segments $P W, Q R, S T$ and $U V$, respectively.



The proof can now be concluded in two similar ways.



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First Approach. Since $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ is a parallelogram, $\overrightarrow{F_{1} F_{2}}+\overrightarrow{F_{3} F_{4}}=\mathbf{0}$ and $\overrightarrow{F_{2} F_{3}}+\overrightarrow{F_{4} F_{1}}=\mathbf{0}$; this still holds in the degenerate case, for if the $O_{i}$ are collinear, then they all lie on the line $T_{1} T_{4}$, and each $O_{i}$ is the midpoint of the segment $T_{i} T_{i+1}$. Consequently,



$$

\begin{aligned}

\overrightarrow{P Q}-\overrightarrow{R S}+\overrightarrow{T U}-\overrightarrow{V W}= & \left(\overrightarrow{P F_{1}}+\overrightarrow{F_{1} F_{2}}+\overrightarrow{F_{2} Q}\right)-\left(\overrightarrow{R F_{2}}+\overrightarrow{F_{2} F_{3}}+\overrightarrow{F_{3} S}\right) \\

& +\left(\overrightarrow{T F_{3}}+\overrightarrow{F_{3} F_{4}}+\overrightarrow{F_{4} U}\right)-\left(\overrightarrow{V F_{4}}+\overrightarrow{F_{4} F_{1}}+\overrightarrow{F_{1} W}\right) \\

= & \left(\overrightarrow{P F_{1}}-\overrightarrow{F_{1} W}\right)-\left(\overrightarrow{R F_{2}}-\overrightarrow{F_{2} Q}\right)+\left(\overrightarrow{T F_{3}}-\overrightarrow{F_{3} S}\right)-\left(\overrightarrow{V F_{4}}-\overrightarrow{F_{4} U}\right) \\

& +\left(\overrightarrow{F_{1} F_{2}}+\overrightarrow{F_{3} F_{4}}\right)-\left(\overrightarrow{F_{2} F_{3}}+\overrightarrow{F_{4} F_{1}}\right)=\mathbf{0} .

\end{aligned}

$$



Alternatively, but equivalently, $\overrightarrow{P Q}+\overrightarrow{T U}=\overrightarrow{R S}+\overrightarrow{V W}$, as required.



Second Approach. This is merely another way of reading the previous argument. Fix an orientation of the line $P W$, say, from $P$ towards $W$, and use a lower case letter to denote the coordinate of a point labelled by the corresponding upper case letter.



Since the diagonals of a parallelogram bisect one another, $f_{1}+f_{3}=f_{2}+f_{4}$, the common value being twice the coordinate of the projection to $P W$ of the point where $O_{1} O_{3}$ and $O_{2} O_{4}$ cross; the relation clearly holds in the degenerate case as well.







Plug $f_{1}=\frac{1}{2}(p+w), f_{2}=\frac{1}{2}(q+r), f_{3}=\frac{1}{2}(s+t)$ and $f_{4}=\frac{1}{2}(u+v)$ into the above equality to get $p+w+s+t=q+r+u+v$. Alternatively, but equivalently, $(q-p)+(u-t)=(s-r)+(w-v)$, that is, $P Q+T U=R Q+V W$, as required.