1963 IMO Problems/Problem 2

Problem

Point $A$ and segment $BC$ are given. Determine the locus of points in space which are the vertices of right angles with one side passing through $A$, and the other side intersecting the segment $BC$.

Solution

Let $\omega_1$ be the circle with diameter $AB$, and let $\omega_2$ be the circle with diameter $AC$. Then the locus is simply the set of points inside either $\omega_1$ or $\omega_2$, but not both.

To see this, suppose the right angle's ray that does not pass through $A$ intersects segment $BC$ at $X$. Then the right angle's vertex must lie on the circle with diameter $AX$. So, for a particular $X$, the desired locus is a circle with diameter $AX$. Accounting for all possible $X$, the total locus is the union of the circumferences of all circles that have a diameter $AX$, where $X$ is some point on $BC$.

As $X$ moves from $B$ to $C$, the motion of the circle with diameter $AX$ is continuous and fluid. Any point $P$ lying within $\omega_1$ but outside $\omega_2$ will eventually be intersected by this moving circle, since it went from inside to outside the circle. This applies similarly to all points inside $\omega_2$ but outside $\omega_1$. Also, the points inside both $\omega_1$ and $\omega_2$ are never intersected by this moving circle, as it always stays inside.

(This proof sucks and needs some formalism)

See Also

1963 IMO (Problems) • Resources)
Preceded by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions
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