Minimum of the perimeter of the triangle

A circle $S$ with arbitrary radius and center is given. Let point $P$ be in the exterior of the circle $S$ and draw the two tangent lines from the point $P$ to the circle $S$. Let the point $A$ be in the circumference of the circle $S$ and let the points $B, C$ be on the two tangent lines. Find the minimum value of the perimeter of the triangle formed by the points $A, B, C$.

My attempt I considered the points $A', A"$ which are formed by the symmetry about the two tangent lines from the point $A$ but can't proceed further. Any help would be appreciated.

• You repeated notation $C$ for the circle and the intersection point of one of the tangents and the circumference. – GNUSupporter 8964民主女神 地下教會 Dec 12 '17 at 6:21
• Thanks, changed it to $S$. – 민찬홍 Dec 12 '17 at 6:23

The solution depends on the fact that shortest lines, like light rays, obey the laws of reflection.

In the diagram below, let $\theta:=\angle OPB$, $\phi=\angle PBC$. If triangle $ABC$ is the optimal triangle, then each corner $ABC$, $BCA$, $CAB$ has equal angles of 'incidence' and of 'reflection' with their respective normals.

Thus, using simple Euclidean theorems, $\angle ABC=2(90^\circ-\phi)$, $\angle BCA=2(\phi+2\theta-90^\circ)$, $\angle BCP=180^\circ-\phi-2\theta$, $\angle CAB = 180^\circ-4\theta$, $\angle AOP=\phi+\theta-90^\circ$. This gives a contradiction unless $\phi+\theta=90^\circ$, that is, $BC$ is perpendicular to $OP$ and $A$ lies on $OP$.

In that case, $ABC$ is isosceles and $\angle PAB=90^\circ-2\theta$, $\angle ABC=\angle BCA=2\theta$ determines its dimensions.

• Why is $\angle AOP=\phi+\theta-90^\circ$? – GNUSupporter 8964民主女神 地下教會 Dec 13 '17 at 16:58
• Let $Q$ be the intersection of $OP$ and $BC$, and $R$ that of $OP$ and $AC$. Then $OQC = 180-\theta-\phi$, so $ORC$ can be found, hence $ORA$ and finally $ROA$. – Chrystomath Dec 13 '17 at 17:37
• I understand $\angle ORC=3\theta+\phi$, but I don't understand how to use this to find $\angle ORA$. Since $A$ is on the circumference, it's related to some circle properties? – GNUSupporter 8964民主女神 地下教會 Dec 13 '17 at 18:21
• Oh I see. You're using the reflection property, and finally you get calculated an angle with negative degree to get a contradiction. – GNUSupporter 8964民主女神 地下教會 Dec 13 '17 at 18:50