[Math] Find the length of the tangent to the circle if the perimeter of the triangle is $8$.

geometry

From the point $O$, two tangents are drawn to a circle such that the perpendicular to one of the tangents at its point of contact with the circle $Q$ meets the other tangent at $P$. If the perimeter of triangle $POQ$ is $8$ units, find the length of tangent to the circle from the point $O$.
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I've tried taking angle $POQ=2\theta$ and applying trigonometry to solve the question but I've not had any success so far.

Best Answer

This problem is undetermined as @Barry Cipra just remarked. Here is why.

Let $OQ=a$ and $QP=b$. The objective is to find $a$.

Perimeter constraint

$$\tag{1}a+b+\sqrt{a^2+b^2}=8$$

can be written in an equivalent way: $(a+b-8)^2=a^2+b^2$ (with condition $a+b<8$ ). Expanding and simplifying, we arrive at the (still equivalent) condition

$$\tag{2}b=8\dfrac{4-a}{8-a}$$

It means that you can choose any $a$, $0 \leq a \leq 4$, there will always exist a unique other possible "leg" side $b$ given by (2), thus there will always exist a right triangle fulfilling condition (1).

Remark:

a) Limit cases $a=0$ and $a=4$ are associated with "flat triangles".

b) A rather simple case is obtained with $a=2$, giving $b=8/3$ and $\sqrt{a^2+b^2}=10/3$. The sides thus obtained are proportionnal to the sides $3,4,5$ of the simplest "pythagorean triple".

Edit: Once you have any of these right triangles, you build the circle:

  • whose center $C$ is at the crossing of the angle bissector of $\widehat{QOP}$ and line $PQ$ and
  • whose radius is $CQ$.
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