Problem 1503

1. Restate the problem in terms of line segments and visibility:
- We need to show that if a set $S$ can be seen from points $A$, $B$, and $C$ of a triangle $ABC$, then $S$ can be seen from any point within the triangle $ABC$.

2. **Consider a point $D$ on the line segment $AB$:**
- We will first show that if $S$ can be seen from $A$ and $B$, then it can be seen from any point $D$ on $AB$.

3. **Choose an arbitrary point $P$ in $S$:**
- Since $S$ can be seen from $A$, every point on the line segment $AP$ is in $S$.
- Similarly, since $S$ can be seen from $B$, every point on the line segment $BP$ is in $S$.

4. **Consider any point $Q$ on the line segment $AP$:**
- Since $Q$ is on $AP$ and $AP \subseteq S$, $Q \in S$.
- Now consider the line segment $BQ$. Since $S$ can be seen from $B$, every point on $BQ$ is in $S$.

5. **Show that $S$ can be seen from $D$:**
- Let $D$ be any point on $AB$. We need to show that every point on the line segment $DP$ is in $S$.
- Since $D$ is on $AB$, it can be expressed as a convex combination of $A$ and $B$, i.e., $D = \lambda A + (1 - \lambda) B$ for some $\lambda \in [0, 1]$.

6. Use the convex combination property:
- Consider any point $R$ on the line segment $DP$. $R$ can be expressed as $R = \mu D + (1 - \mu) P$ for some $\mu \in [0, 1]$.
- Substituting $D = \lambda A + (1 - \lambda) B$ into the expression for $R$, we get:
$$R = \mu (\lambda A + (1 - \lambda) B) + (1 - \mu) P$$
$$R = \mu \lambda A + \mu (1 - \lambda) B + (1 - \mu) P$$
- Since $A$, $B$, and $P$ are all points from which $S$ can be seen, and $S$ is convex, $R$ must also be in $S$.

7. **Generalize to any point within the triangle $ABC$:**
- By the same argument, any point on the line segment $CD$ can see $S$ because $S$ can be seen from $C$ and $D$.
- Therefore, any point within the triangle $ABC$ can see $S$ because it can be expressed as a convex combination of points from which $S$ can be seen.

$\blacksquare$