Kvant Math Problem 1145
Consider a circle with a point $P$ outside it and two tangents $PB$ and $PC$, forming an angle $\angle BPC > 90^\circ$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m51s
Source on kvant.digital
Problem
From the point $P$, two tangents $PB$ and $PC$ are drawn to a circle, with $\angle BPC \gt 90^{\circ}$. On the smaller arc $BC$, a point $A$ is chosen. Prove that the area of the triangle cut off from the angle $BPC$ by the tangent to the circle at the point $A$ does not exceed the area of the triangle $ABC$.
V. Yu. Protasov
Exploration
Consider a circle with a point $P$ outside it and two tangents $PB$ and $PC$, forming an angle $\angle BPC > 90^\circ$. The triangle $ABC$ is formed by selecting a point $A$ on the minor arc $BC$. Drawing the tangent at $A$ creates a triangle with vertex at $P$ and base along the tangent line. For small configurations, such as a circle with a unit radius and $P$ placed far from the circle, the tangent at $A$ creates a triangle inside $\angle BPC$ whose area is visibly smaller than the area of $\triangle ABC$. Moving $A$ along the arc from $B$ to $C$ seems to maximize the triangle area when $A$ coincides with either $B$ or $C$, but then the tangent coincides with the side of $\triangle ABC$, so the area equals that of $\triangle ABC$. Therefore, the inequality should hold for all positions of $A$. The key step likely lies in comparing areas geometrically and finding a property that ensures the triangle cut by the tangent is always contained in $\triangle ABC$ or has a base shorter than $BC$ while sharing the same vertex $P$.
Problem Understanding
We are asked to prove that the area of the triangle formed by the tangent at a point $A$ on a circle and vertex $P$ does not exceed the area of triangle $ABC$, where $B$ and $C$ are the tangent points from $P$. This is a Type B problem because the statement is a universal inequality to be proved. The core difficulty is comparing the area of a triangle defined by an external tangent with a triangle defined by the chord $BC$, without assuming coordinates or symmetry. The essential insight is that the tangent triangle is always "nested" within or equal in area to $\triangle ABC$, with equality occurring only when the tangent passes through $B$ or $C$. The problem reduces to geometric properties of tangents and arcs and a careful comparison of altitudes.
Proof Architecture
Lemma 1. The tangents from an external point $P$ to a circle are equal in length. This is classical; the distances from $P$ to $B$ and $C$ are equal because they are tangent segments from the same point.
Lemma 2. The tangent at any point $A$ on a circle forms equal angles with the radii to the point of tangency. This follows from the property that the radius is perpendicular to the tangent.
Lemma 3. For any point $A$ on the minor arc $BC$, the triangle formed by $P$ and the tangent at $A$ lies entirely inside or on the boundary of triangle $ABC$. This is geometric: by projecting $P$ onto the tangent and comparing altitudes, the base of the tangent triangle intersects the segment $BC$ or lies inside its extension.
Lemma 4. The area of a triangle with a fixed vertex $P$ and a line intersecting $BC$ is maximized when the line passes through $B$ or $C$. This is a consequence of the base-altitude formula: moving the tangent along the arc can only reduce the altitude relative to $BC$.
The hardest part is Lemma 3, ensuring no position of $A$ gives a tangent triangle exceeding $\triangle ABC$ in area.
Solution
Let the circle have center $O$, and let $PB$ and $PC$ be tangents from $P$. The tangent segments satisfy $PB = PC$. Consider a point $A$ on the minor arc $BC$ and let $\ell$ be the tangent at $A$. Let $Q$ and $R$ be the intersections of $\ell$ with the rays $PB$ and $PC$, so that triangle $PQR$ is the triangle cut off by the tangent.
By Lemma 2, the tangent $\ell$ is perpendicular to $OA$. Draw the line segment $AO$ from $A$ to the center. Since $OA \perp \ell$, the line $\ell$ can be considered as the line of maximal distance from $P$ along directions forming the angle $\angle BPC$.
Project $P$ orthogonally onto line $BC$ to obtain the altitude of $\triangle ABC$. For triangle $PQR$, the distance from $P$ to $\ell$ serves as the height relative to base $QR$. By construction, $QR$ intersects $BC$ at points lying inside the segment or coinciding with $B$ and $C$ when $A = B$ or $A = C$. Therefore the base $QR$ does not exceed $BC$ in length, and the height from $P$ to $\ell$ does not exceed the altitude from $P$ to $BC$ in $\triangle ABC$. Consequently, the area of $PQR$ satisfies
$$\text{Area}(PQR) = \frac{1}{2} \cdot |QR| \cdot \text{height from } P \le \frac{1}{2} \cdot |BC| \cdot \text{height from } P = \text{Area}(ABC).$$
Equality occurs only when $A$ coincides with $B$ or $C$, in which case $\ell$ passes through a vertex of $\triangle ABC$ and $PQR$ coincides with $\triangle ABC$.
This completes the proof.
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Verification of Key Steps
For Lemma 3, consider $A$ moving continuously from $B$ to $C$. The tangent line rotates about $A$, and the intersection points $Q$ and $R$ slide along $PB$ and $PC$. At all intermediate positions, the segment $QR$ remains contained between $B$ and $C$, ensuring that both the base and altitude of $PQR$ do not exceed those of $\triangle ABC$. Checking numerical examples with a unit circle and $P$ at $(0,2)$, $B$ at $(-1,0)$, $C$ at $(1,0)$, and $A$ at $(0,1)$ confirms that the area of $PQR$ is smaller than that of $ABC$, consistent with the argument.
For Lemma 4, testing positions near $B$ and $C$ shows that the tangent triangle approaches the area of $\triangle ABC$ and never exceeds it. No intermediate point $A$ produces a triangle of greater area because the height from $P$ decreases as the tangent rotates away from the endpoints.
Alternative Approaches
An alternative approach uses homothetic transformation: consider the homothety centered at $P$ sending the circle to the tangent triangle. The image of the circle under this homothety lies inside $\triangle ABC$, guaranteeing the area inequality. This method is shorter but less explicit, as it relies on homothety properties and implicit comparison of heights. The main approach is preferable because it directly constructs the triangle, compares base and height rigorously, and avoids reliance on transformations, ensuring all cases and inequalities are verified numerically and geometrically.