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.

  • $\begingroup$ You repeated notation $C$ for the circle and the intersection point of one of the tangents and the circumference. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 12 '17 at 6:21
  • $\begingroup$ Thanks, changed it to $S$. $\endgroup$ – 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 $\theta:=\angle OPB$, $\phi=\angle PBC$. enter image description here

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.

  • $\begingroup$ Why is $\angle AOP=\phi+\theta-90^\circ$? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 13 '17 at 16:58
  • $\begingroup$ 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 $ROA$. $\endgroup$ – Chrystomath Dec 13 '17 at 17:37
  • $\begingroup$ 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? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 13 '17 at 18:21
  • $\begingroup$ Oh I see. You're using the reflection property, and finally you get calculated an angle with negative degree to get a contradiction. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 13 '17 at 18:50

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