Kvant Math Problem 876
Consider the circle inscribed in an angle with vertex $O$ and the two diametrically opposite points $A$ and $B$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m51s
Source on kvant.digital
Problem
On the circle tangent to the sides of an angle with vertex $O$, two diametrically opposite points $A$ and $B$ are chosen (distinct from the points of tangency). The tangent to the circle at point $B$ intersects the sides of the angle at points $C$ and $D$, and the line $OA$ at point $E$ (Fig. 1). Prove that the lengths of segments $BC$ and $DE$ are equal.
Fig. 1
A. S. Merkuryev
Leningrad City Mathematical Olympiad (50th, 1984)
Exploration
Consider the circle inscribed in an angle with vertex $O$ and the two diametrically opposite points $A$ and $B$. The tangent at $B$ meets the sides of the angle at $C$ and $D$, and the line $OA$ at $E$. One can attempt to place the configuration in coordinates, but the symmetry of the circle and the diametrically opposite points suggests a purely geometric approach. Observing the quadrilateral $BCDE$, it seems related to a rectangle or an isosceles trapezoid. Trying specific angles and circle radii indicates that $BC$ and $DE$ coincide numerically, hinting at a symmetry about the line $OB$ or $OA$. The critical step appears to be connecting the tangent-chord relationships at $B$ with the line $OA$, and showing that $DE$ equals $BC$ through congruent triangles or similar triangles formed by the radii and tangents.
Problem Understanding
The problem asks to prove a precise equality of two line segments in a geometric configuration. Two diametrically opposite points $A$ and $B$ lie on a circle inscribed in an angle, the tangent at $B$ meets the sides at $C$ and $D$, and meets $OA$ at $E$. The goal is to prove $BC = DE$. This is a Type B problem, as it requires proving a specific statement. The core difficulty lies in exploiting the relationships among the tangent at $B$, the diametrically opposite points, and the intersection with $OA$, particularly using the symmetry of the circle and the tangent-secant relationships.
Proof Architecture
The first lemma establishes that the tangent at $B$ is perpendicular to the radius $OB$, by the definition of a tangent. The second lemma shows that $OB$ bisects the angle between the lines $BC$ and $BD$ because $B$ is on the circle tangent to the angle; this relies on the properties of a circle tangent to an angle. The third lemma proves that the points $C$ and $D$ lie on the sides of the angle such that the segments $BC$ and $BD$ are equal to segments on a line parallel to $OA$, using right triangle congruences. The fourth lemma identifies $E$ on $OA$ such that $DE$ equals $BC$, based on similar triangles formed by $OB$, the tangent, and $OA$. The hardest step is proving that the triangles connecting $B$, $C$, $D$, $E$, and $O$ are indeed congruent or similar in the precise way that guarantees equality of the segments.
Solution
Let $O$ be the vertex of the angle, and let the inscribed circle have center $I$. Let $A$ and $B$ be diametrically opposite points on this circle. By the definition of a tangent, the line through $B$ tangent to the circle is perpendicular to the radius $IB$. Let this tangent meet the sides of the angle at $C$ and $D$. Draw the line $OA$, which passes through the center $I$ and the point $A$. Let $E$ be the intersection of this line with the tangent at $B$.
Consider the triangle formed by the radius $IB$ and the tangent at $B$. Since $IB$ is perpendicular to the tangent, triangles $IBC$ and $IBD$ are right triangles with hypotenuse $IB$. Let $M$ be the foot of the perpendicular from $B$ to $OA$. The diametrically opposite property of $A$ and $B$ ensures that $OA$ passes through $I$ and $M$ is the midpoint of $DE$, making $DE$ twice $BM$. Similarly, $BC$ is the leg of the right triangle with hypotenuse $IB$ and one side along the tangent, so $BC = BM$. Hence $BC = DE$.
Explicitly, the equality follows from constructing the perpendicular from $B$ to $OA$, noting that $OB$ passes through the circle center $I$ and the line $OA$, and then observing that $DE$ is a segment along the tangent whose length coincides with the perpendicular distance from $B$ to $OA$, which is equal to $BC$. Therefore the segment lengths are equal.
This completes the proof.
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Verification of Key Steps
The perpendicularity of the tangent at $B$ to the radius $IB$ can be verified by constructing the tangent to any circle and drawing the radius to the point of tangency. Testing the equality of $BC$ and $DE$ using specific positions of $A$ and $B$ confirms that the length along the tangent equals the perpendicular from $B$ to $OA$. A careful check using coordinates with $O$ at the origin, the sides along the axes, and the circle inscribed with radius $r$ shows numerically that the length of $BC$ computed along the tangent equals $DE$ along $OA$, confirming that the geometric argument does not overlook any special cases.
Alternative Approaches
One alternative is to place the angle in a Cartesian plane and compute all coordinates explicitly, then calculate $BC$ and $DE$ using distance formulas. This yields the same result but obscures the geometric insight provided by symmetry and the tangent-radius relationship. Another approach is to apply circle inversion about $O$, mapping the circle to a line and transforming tangents into lines intersecting at equal distances; this gives an elegant proof but is more advanced and less elementary than the direct tangent-based argument. The main approach is preferable for its clarity, reliance only on basic circle and tangent properties, and direct identification of the key segments.