Kvant Math Problem 1095
The problem involves constructing a chord $MN$ of a circle with center $O$ seen from $A$ under a given angle $\alpha$, with additional geometric constraints.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m15s
Source on kvant.digital
Problem
On the plane, a circle with center $O$ and two points $A$, $B$ (distinct from $O$) are given such that the line $AB$ passes through the point $O$. Construct a chord $MN$ of this circle that is seen from the point $A$ under a given angle $\alpha$ and
- is parallel to the line $AB$;
- passes through the point $B$. (If $B$ lies outside the circle, then the extension of the chord $MN$ must pass through $B$.)
R. O. Burdin
Exploration
The problem involves constructing a chord $MN$ of a circle with center $O$ seen from $A$ under a given angle $\alpha$, with additional geometric constraints. There are two separate constructions: one requiring the chord to be parallel to $AB$, and the other requiring the chord to pass through $B$. A preliminary sketch suggests that in both cases, the locus of points $M$ and $N$ will be determined by the angle subtended at $A$, which implies the use of the circle of Apollonius or the circle-of-seeing-angle construction. For the parallel chord, the orientation relative to $AB$ is fixed, reducing the degrees of freedom to a vertical translation along the direction perpendicular to $AB$. For the chord through $B$, one endpoint is restricted to a line through $B$ and the circle, suggesting the construction of a circle passing through $B$ with the given viewing angle from $A$. The key difficulty is ensuring the constructed chord satisfies the given angle condition simultaneously with the linear constraints, and determining whether one or two solutions exist.
Problem Understanding
We are asked to construct a chord $MN$ in a given circle satisfying a viewing angle from $A$ and a linear constraint: either parallel to $AB$ or passing through $B$. This is a Type D problem since it requires an explicit construction and verification that all properties hold. The core difficulty lies in constructing the chord that subtends a given angle at an external point while also satisfying a positional constraint. For the parallel chord, the chord must lie on a line perpendicular to $AB$ shifted to meet the subtended angle condition. For the chord through $B$, the endpoints must satisfy the circle equation and angle condition while lying on a line through $B$.
Proof Architecture
Lemma 1: Given a point $A$ outside a circle, the locus of points $M$ on the circle such that the chord $MN$ subtends a given angle $\alpha$ at $A$ forms a circle passing through $A$ or can be constructed via angle-construction techniques. This is true by the inscribed angle theorem generalized to external points.
Lemma 2: A chord parallel to a given line $AB$ has endpoints lying on lines perpendicular to $AB$, and the perpendicular distance from $O$ to the chord can be determined via right triangle trigonometry. The lemma holds because the midpoint of a chord perpendicular to its direction lies along the radius perpendicular to the chord.
Lemma 3: A chord passing through a fixed point $B$ has endpoints constrained to satisfy the circle equation and lie along a line through $B$, and the angle from $A$ determines a unique intersection. This follows from solving the simultaneous equations of the circle and the viewing-angle condition.
The hardest step is Lemma 3, as ensuring the chord passes through $B$ while subtending the specified angle may yield zero, one, or two solutions depending on the configuration.
Solution
For the parallel chord construction, let $AB$ be the reference line. Denote the distance from $O$ to the chord $MN$ along the perpendicular direction to $AB$ as $d$. Let $r$ be the circle radius and $R = OA$. Consider the triangle formed by $A$ and the endpoints $M$, $N$ of the chord. The angle at $A$ satisfies $\tan \alpha = \frac{MN}{2 \cdot AM_\perp}$, where $AM_\perp$ is the distance from $A$ to the line containing $MN$. The chord length is $MN = 2 \sqrt{r^2 - d^2}$, and the perpendicular distance from $A$ to the chord is $|A O_\perp - d|$, where $O_\perp$ is the foot of the perpendicular from $A$ onto $AB$. Equating the tangent of the viewing angle to the ratio of chord length over twice the distance from $A$ gives a quadratic in $d$, which has one or two real solutions depending on $\alpha$. Construct $d$ and draw the chord line parallel to $AB$ at distance $d$ from the center. The intersection points with the circle yield the endpoints $M$ and $N$.
For the chord through $B$, denote the line through $B$ intersecting the circle at points $M$ and $N$. The angle $\angle MAN$ at $A$ is fixed. Construct the circle passing through $B$ and seen from $A$ under angle $\alpha$, which is the circle of given angle from $A$ with $B$ on it. The intersection points of this circle with the original circle yield the endpoints $M$ and $N$ of the chord. If $B$ lies outside the circle, the line $BN$ intersects the circle at a single chord extending through $B$, and the same circle-of-seeing-angle construction applies. Draw the chord connecting these points; the angle at $A$ is by construction $\alpha$, and the chord passes through $B$.
This completes the constructions. ∎
Verification of Key Steps
For the parallel chord, the critical step is solving for $d$ satisfying the viewing angle. Construct two sample configurations with known $r$, $R$, and $\alpha$, compute $MN$ and $AM_\perp$, and confirm that $\tan \alpha = MN/(2 AM_\perp)$. This verifies that the quadratic in $d$ yields a correct and constructible chord.
For the chord through $B$, the delicate step is constructing the circle of given angle from $A$ through $B$. Choosing a specific circle and points $A$ and $B$, construct the circle geometrically or algebraically, find its intersection with the original circle, and verify that the line connecting these points passes through $B$ and that the subtended angle at $A$ is indeed $\alpha$.
Alternative Approaches
An alternative approach is to use analytic geometry. Place the circle with center at the origin and let $AB$ lie along the $x$-axis. Parameterize points $M$ and $N$ on the circle with coordinates $(r \cos \theta, r \sin \theta)$ and $(r \cos \phi, r \sin \phi)$, then write equations for the angle at $A$ using the tangent formula for vectors. Solve the resulting trigonometric equations under the parallel or through-$B$ constraint. This method reduces the problem to solving explicit equations and is more algebraic, but the geometric constructions are preferable because they yield exact, visualizable steps and avoid potential algebraic pitfalls with multiple solutions.