limiting ratio of two triangle areas

Question

At the end points and the midpoint of circular arc $AB$ tangent lines are drawn, and the points $A$ and $B$ are joined with a chord. Find the limit of the ratio of the areas of the two triangles thus formed, as the length of arc $AB$ decreases indefinitely, approaching a length of zero.

Answer 1

Without loss of generality, we can assume the circle has radius $1$.

Let $O$ be the center of the circle.

Let $C$ be the point where the tangent lines at $A$ and $B$ intersect.

Let $M$ be the midpoint of segment $AB$, and let $N$ be the midpoint of arc $AB$.

Since we are intending to force the length of arc $AB$ to approach zero, we can assume that angle $AOB$ is acute.

Let $A',B'$ be the points where the tangent line at $N$ meets $AC,BC$, respectively.

Let $\theta=\angle AOC$.

Then $\angle AOB = 2\theta$.

Our goal is to find the limit of the ratio $$ {\frac {\text{area}(\Delta A'B'C)} {\text{area}(\Delta ABC)} } $$ as $2\theta$ approaches $0$, or equivalently, as $\theta$ approaches $0$.

By right triangle trigonometry, we get

• $CO=\sec(\theta)$, hence $CN=\sec(\theta)-1$.$\\[4pt]$
• $MO=\cos(\theta)$, hence $CM=\sec(\theta)-cos(\theta)$.

Since triangles $A'B'C$ and $ABC$ are similar, with corresponding altitudes $CN$ and $CM$, respectively, it follows that $$ {\frac {\text{area}(\Delta A'B'C)} {\text{area}(\Delta ABC)} } = \left(\!\frac{CN}{CM}\!\right)^{\!2} = \left(\!\frac{\sec(\theta)-1}{\sec(\theta)-\cos(\theta)}\!\right)^{\!2} $$ which simplifies to $$ \left(\!\frac{1}{1+\cos(\theta)}\!\right)^{\!2} $$ hence the limit, as $\theta$ approaches $0$, is ${\large{\frac{1}{4}}}$.