exam: RMM

problem: Let $x$ and $y$ be positive real numbers such that $x+y^{2016} \geq 1$. Prove that $x^{2016}+y>$ 1 - 1/100.

solution:
If $x \geq 1-1 /(100 \cdot 2016)$, then

$$
x^{2016} \geq\left(1-\frac{1}{100 \cdot 2016}\right)^{2016}>1-2016 \cdot \frac{1}{100 \cdot 2016}=1-\frac{1}{100}
$$

by Bernoulli's inequality, whence the conclusion.
If $x<1-1 /(100 \cdot 2016)$, then $y \geq(1-x)^{1 / 2016}>(100 \cdot 2016)^{-1 / 2016}$, and it is sufficient to show that the latter is greater than $1-1 / 100=99 / 100$; alternatively, but equivalently, that

$$
\left(1+\frac{1}{99}\right)^{2016}>100 \cdot 2016
$$

To establish the latter, refer again to Bernoulli's inequality to write

$$
\left(1+\frac{1}{99}\right)^{2016}>\left(1+\frac{1}{99}\right)^{99 \cdot 20}>\left(1+99 \cdot \frac{1}{99}\right)^{20}=2^{20}>100 \cdot 2016
$$

Remarks. (1) Although the constant $1 / 100$ is not sharp, it cannot be replaced by the smaller constant $1 / 400$, as the values $x=1-1 / 210$ and $y=1-1 / 380$ show.
(2) It is natural to ask whether $x^{n}+y \geq 1-1 / k$, whenever $x$ and $y$ are positive real numbers such that $x+y^{n} \geq 1$, and $k$ and $n$ are large. Using the inequality $\left(1+\frac{1}{k-1}\right)^{k}>\mathrm{e}$, it can be shown along the lines in the solution that this is indeed the case if $k \leq \frac{n}{2 \log n}(1+o(1))$. It seems that this estimate differs from the best one by a constant factor.

problem_match: Problem 4.

solution_match: Solution.


problem:
A convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$ is inscribed in a circle $\Omega$ of radius $R$. The diagonals $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ concur at $X$. For $i=1,2,3$, let $\omega_{i}$ be the circle tangent to the segments $X A_{i}$ and $X B_{i}$, and to the arc $A_{i} B_{i}$ of $\Omega$ not containing other vertices of the hexagon; let $r_{i}$ be the radius of $\omega_{i}$.
(a) Prove that $R \geq r_{1}+r_{2}+r_{3}$.
(b) If $R=r_{1}+r_{2}+r_{3}$, prove that the six points where the circles $\omega_{i}$ touch the diagonals $A_{1} B_{2}, A_{2} B_{3}, A_{3} B_{1}$ are concyclic.

solution:
(a) Let $\ell_{1}$ be the tangent to $\Omega$ parallel to $A_{2} B_{3}$, lying on the same side of $A_{2} B_{3}$ as $\omega_{1}$. The tangents $\ell_{2}$ and $\ell_{3}$ are defined similarly. The lines $\ell_{1}$ and $\ell_{2}, \ell_{2}$ and $\ell_{3}, \ell_{3}$ and $\ell_{1}$ meet at $C_{3}, C_{1}, C_{2}$, respectively (see Fig. 1). Finally, the line $C_{2} C_{3}$ meets the rays $X A_{1}$ and $X B_{1}$ emanating from $X$ at $S_{1}$ and $T_{1}$, respectively; the points $S_{2}, T_{2}$, and $S_{3}, T_{3}$ are defined similarly.

Each of the triangles $\Delta_{1}=\triangle X S_{1} T_{1}, \Delta_{2}=\triangle T_{2} X S_{2}$, and $\Delta_{3}=\triangle S_{3} T_{3} X$ is similar to $\Delta=\triangle C_{1} C_{2} C_{3}$, since their corresponding sides are parallel. Let $k_{i}$ be the ratio of similitude of $\Delta_{i}$ and $\Delta$ (e.g., $k_{1}=X S_{1} / C_{1} C_{2}$ and the like). Since $S_{1} X=C_{2} T_{3}$ and $X T_{2}=S_{3} C_{1}$, it follows that $k_{1}+k_{2}+k_{3}=1$, so, if $\rho_{i}$ is the inradius of $\Delta_{i}$, then $\rho_{1}+\rho_{2}+\rho_{3}=R$.

Finally, notice that $\omega_{i}$ is interior to $\Delta_{i}$, so $r_{i} \leq \rho_{i}$, and the conclusion follows by the preceding.
![](https://cdn.mathpix.com/cropped/2024_11_22_43c8c12599eee71f4e24g-2.jpg?height=712&width=1054&top_left_y=181&top_left_x=221)

Fig. 1
![](https://cdn.mathpix.com/cropped/2024_11_22_43c8c12599eee71f4e24g-2.jpg?height=649&width=492&top_left_y=241&top_left_x=1290)

Fig. 2
(b) By part (a), the equality $R=r_{1}+r_{2}+r_{3}$ holds if and only if $r_{i}=\rho_{i}$ for all $i$, which implies in turn that $\omega_{i}$ is the incircle of $\Delta_{i}$. Let $K_{i}, L_{i}, M_{i}$ be the points where $\omega_{i}$ touches the sides $X S_{i}, X T_{i}, S_{i} T_{i}$, respectively. We claim that the six points $K_{i}$ and $L_{i}(i=1,2,3)$ are equidistant from $X$.

Clearly, $X K_{i}=X L_{i}$, and we are to prove that $X K_{2}=X L_{1}$ and $X K_{3}=X L_{2}$. By similarity, $\angle T_{1} M_{1} L_{1}=\angle C_{3} M_{1} M_{2}$ and $\angle S_{2} M_{2} K_{2}=\angle C_{3} M_{2} M_{1}$, so the points $M_{1}, M_{2}, L_{1}, K_{2}$ are collinear. Consequently, $\angle X K_{2} L_{1}=\angle C_{3} M_{1} M_{2}=\angle C_{3} M_{2} M_{1}=\angle X L_{1} K_{2}$, so $X K_{2}=X L_{1}$. Similarly, $X K_{3}=X L_{2}$.

Remark. Under the assumption in part (b), the point $M_{i}$ is the centre of a homothety mapping $\omega_{i}$ to $\Omega$. Since this homothety maps $X$ to $C_{i}$, the points $M_{i}, C_{i}, X$ are collinear, so $X$ is the Gergonne point of the triangle $C_{1} C_{2} C_{3}$. This condition is in fact equivalent to $R=r_{1}+r_{2}+r_{3}$.

problem_match: Problem 5.


problem: A set of $n$ points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets $\mathcal{A}$ and $\mathcal{B}$. An $\mathcal{A B}$-tree is a configuration of $n-1$ segments, each of which has an endpoint in $\mathcal{A}$ and the other in $\mathcal{B}$, and such that no segments form a closed polyline. An $\mathcal{A B}$-tree is transformed into another as follows: choose three distinct segments $A_{1} B_{1}, B_{1} A_{2}$ and $A_{2} B_{2}$ in the $\mathcal{A B}$-tree such that $A_{1}$ is in $\mathcal{A}$ and $A_{1} B_{1}+A_{2} B_{2}>A_{1} B_{2}+A_{2} B_{1}$, and remove the segment $A_{1} B_{1}$ to replace it by the segment $A_{1} B_{2}$. Given any $\mathcal{A B}$-tree, prove that every sequence of successive transformations comes to an end (no further transformation is possible) after finitely many steps.

solution:
The configurations of segments under consideration are all bipartite geometric trees on the points $n$ whose vertex-parts are $\mathcal{A}$ and $\mathcal{B}$, and transforming one into another preserves the degree of any vertex in $\mathcal{A}$, but not necessarily that of a vertex in $\mathcal{B}$.

The idea is to devise a strict semi-invariant of the process, i.e., assign each $\mathcal{A B}$-tree a real number strictly decreasing under a transformation. Since the number of trees on the $n$ points is finite, the conclusion follows.

To describe the assignment, consider an $\mathcal{A B}$-tree $\mathcal{T}=(\mathcal{A} \sqcup \mathcal{B}, \mathcal{E})$. Removal of an edge $e$ of $\mathcal{T}$ splits the graph into exactly two components. Let $p_{\mathcal{T}}(e)$ be the number of vertices in $\mathcal{A}$ lying in the component of $\mathcal{T}-e$ containing the $\mathcal{A}$-endpoint of $e$; since $\mathcal{T}$ is a tree, $p_{\mathcal{T}}(e)$ counts the number of paths in $\mathcal{T}-e$ from the $\mathcal{A}$-endpoint of $e$ to vertices in $\mathcal{A}$ (including the one-vertex path). Define $f(\mathcal{T})=\sum_{e \in \mathcal{E}} p_{\mathcal{T}}(e)|e|$, where $|e|$ is the Euclidean length of $e$.

We claim that $f$ strictly decreases under a transformation. To prove this, let $\mathcal{T}^{\prime}$ be obtained from $\mathcal{T}$ by a transformation involving the polyline $A_{1} B_{1} A_{2} B_{2}$; that is, $A_{1}$ and $A_{2}$ are in $\mathcal{A}, B_{1}$
and $B_{2}$ are in $\mathcal{B}, A_{1} B_{1}+A_{2} B_{2}>A_{1} B_{2}+A_{2} B_{1}$, and $\mathcal{T}^{\prime}=\mathcal{T}-A_{1} B_{1}+A_{1} B_{2}$. It is readily checked that $p_{\mathcal{T}^{\prime}}(e)=p_{\mathcal{T}}(e)$ for every edge $e$ of $\mathcal{T}$ different from $A_{1} B_{1}, A_{2} B_{1}$ and $A_{2} B_{2}, p_{\mathcal{T}^{\prime}}\left(A_{1} B_{2}\right)=$ $p_{\mathcal{T}}\left(A_{1} B_{1}\right), p_{\mathcal{T}^{\prime}}\left(A_{2} B_{1}\right)=p_{\mathcal{T}}\left(A_{2} B_{1}\right)+p_{\mathcal{T}}\left(A_{1} B_{1}\right)$, and $p_{\mathcal{T}^{\prime}}\left(A_{2} B b_{2}\right)=p_{\mathcal{T}}\left(A_{2} B_{2}\right)-p_{\mathcal{T}}\left(A_{1} B_{1}\right)$. Consequently,

$$
\begin{aligned}
f\left(\mathcal{T}^{\prime}\right)-f(\mathcal{T})= & p_{\mathcal{T}^{\prime}}\left(A_{1} B_{2}\right) \cdot A_{1} B_{2}+\left(p_{\mathcal{T}^{\prime}}\left(A_{2} B_{1}\right)-p_{\mathcal{T}}\left(A_{2} B_{1}\right)\right) \cdot A_{2} B_{1}+ \\
& \left(p_{\mathcal{T}^{\prime}}\left(A_{2} B_{2}\right)-p_{\mathcal{T}}\left(A_{2} B_{2}\right)\right) \cdot A_{2} B_{2}-p_{\mathcal{T}}\left(A_{1} B_{1}\right) \cdot A_{1} B_{1} \\
= & p_{\mathcal{T}}\left(A_{1} B_{1}\right)\left(A_{1} B_{2}+A_{2} B_{1}-A_{2} B_{2}-A_{1} B_{1}\right)<0
\end{aligned}
$$

Remarks. (1) The solution above does not involve the geometric structure of the configurations, so the conclusion still holds if the Euclidean length (distance) is replaced by any real-valued function on $\mathcal{A} \times \mathcal{B}$.
(2) There are infinitely many strict semi-invariants that can be used to establish the conclusion, as we are presently going to show. The idea is to devise a non-strict real-valued semiinvariant $f_{A}$ for each $A$ in $\mathcal{A}$ (i.e., $f_{A}$ does not increase under a transformation) such that $\sum_{A \in \mathcal{A}} f_{A}=f$. It then follows that any linear combination of the $f_{A}$ with positive coefficients is a strict semi-invariant.

To describe $f_{A}$, where $A$ is a fixed vertex in $\mathcal{A}$, let $\mathcal{T}$ be an $\mathcal{A B}$-tree. Since $\mathcal{T}$ is a tree, by orienting all paths in $\mathcal{T}$ with an endpoint at $A$ away from $A$, every edge of $\mathcal{T}$ comes out with a unique orientation so that the in-degree of every vertex of $\mathcal{T}$ other than $A$ is 1 . Define $f_{A}(\mathcal{T})$ to be the sum of the Euclidean lengths of all out-going edges from $\mathcal{A}$. It can be shown that $f_{A}$ does not increase under a transformation, and it strictly decreases if the paths from $A$ to each of $A_{1}$, $A_{2}, B_{1}, B_{2}$ all pass through $A_{1}$ - i.e., of these four vertices, $A_{1}$ is combinatorially nearest to $A$. In particular, this is the case if $A_{1}=A$, i.e., the edge-switch in the transformation occurs at $A$. It is not hard to prove that $\sum_{A \in \mathcal{A}} f_{A}(\mathcal{T})=f(\mathcal{T})$.

The conclusion of the problem can also be established by resorting to a single carefully chosen $f_{A}$. Suppose, if possible, that the process is infinite, so some tree $\mathcal{T}$ occurs (at least) twice. Let $A$ be the vertex in $\mathcal{A}$ at which the edge-switch occurs in the transformation of the first occurrence of $\mathcal{T}$. By the preceding paragraph, consideration of $f_{A}$ shows that $\mathcal{T}$ can never occur again.
(3) Recall that the degree of any vertex in $\mathcal{A}$ is invariant under a transformation, so the linear combination $\sum_{A \in \mathcal{A}}(\operatorname{deg} A-1) f_{A}$ is a strict semi-invariant for $\mathcal{A B}$-trees $\mathcal{T}$ whose vertices in $\mathcal{A}$ all have degrees exceeding 1. Up to a factor, this semi-invariant can alternatively, but equivalently be described as follows. Fix a vertex $*$ and assign each vertex $X$ a number $g(X)$ so that $g(*)=0$, and $g(A)-g(B)=A B$ for every $A$ in $\mathcal{A}$ and every $B$ in $\mathcal{B}$ joined by an edge. Next, let $\beta(\mathcal{T})=\frac{1}{|\mathcal{B}|} \sum_{B \in \mathcal{B}} g(B)$, let $\alpha(\mathcal{T})=\frac{1}{|\mathcal{E}|-|\mathcal{A}|} \sum_{A \in \mathcal{A}}(\operatorname{deg} A-1) g(A)$, where $\mathcal{E}$ is the edge-set of $\mathcal{T}$, and set $\mu(\mathcal{T})=\beta(\mathcal{T})-\alpha(\mathcal{T})$. It can be shown that $\mu$ strictly decreases under a transformation; in fact, $\mu$ and $\sum_{A \in \mathcal{A}}(\operatorname{deg} A-1) f_{A}$ are proportional to one another.

problem_match: Problem 6.