Problem 7. (Equal Perimeter Problem)

Consider the circle drawn as below with a point C outside the circle.

Now draw from C the two tangent lines to the circle, touching the circle at P₁ and P₂. Now draw a line AB connecting the tangent lines CP₁ and CP₂ which is itself tangent to the circle. Show that the perimeter (distance around) of the resulting triangle ABC is the same no matter how the line AB connecting the tangent lines is drawn.

Solution

We first realize that the two tangent lines CP₁ and CP₂ have the same length, and that the perimeter of the triangle ABC is AB + BC + CA. But if we draw the point D where the line AB touches the circle and realize that the triangles OP₂ A and ODA are congruent, as are the triangles OP₁ B and ODB, we have equality of the distance P₁ B = DB and P₂ A = BA. See the figure below. Hence,

AB = P₁ A + P₂ A

and so the perimeter of the triangle ABC is

AB + BC + CA = CP₁ + CP₂

Hence, the perimeter of the triangle ABC is the same no matter now the line AB is drawn.

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Last modified on Wednesday, March 17, 1999