The circle $A$ touches the circle $B$ internally at $P$. The centre $O$ of $B$ is outside $A$. Let $XY$ be a diameter of $B$ which is also tangent to $A$. Assume $PY > PX$. Let $PY$ intersect $A$ at $Z$. If $Y Z = 2PZ$, what is the magnitude of $\angle PYX$ in degrees?

What I have tried:

  1. Obviously, the red angles are equal, and the orange angles are equal. This gives $XY \parallel TZ$.
  2. $YZ=2PZ$. From this $XY=3TZ$ then $O'Z=3OY$. Let $O'Z=a=O'S$ so $SZ=\sqrt{2} a$, and also $O'O=2a$
  3. Then $SO=\sqrt{3} a$. Now we can use trigonometry to find $\angle PYX$ in triangle $ZSY$.


Please verify whether my figure is correct. Your solution to this question is welcomed, especially if it is shorter.

  • $\begingroup$ Why do we have $SO=3O'Z$? From $3TZ=XY$ it follows that $XO=3O'Z$, so we can't possible have $SO=3O'Z$... $\endgroup$
    – math635
    Commented Dec 23, 2015 at 4:40
  • $\begingroup$ Yah I have edited now $\endgroup$
    – mnulb
    Commented Dec 23, 2015 at 4:44
  • $\begingroup$ It's wrong again. Do check what you write before you ask other people. $\endgroup$
    – user21820
    Commented Dec 23, 2015 at 4:54
  • $\begingroup$ Your solution looks more or less correct to me, but i have to concerns. 1. How do you find $ZY$ in terms of $a$? 2. How can you use trigonometry to determine$\angle PYX$ $\endgroup$
    – math635
    Commented Dec 23, 2015 at 5:02
  • $\begingroup$ we can use power of point with Y i.e. $YS^2%=Yz*YP. $\endgroup$
    – mnulb
    Commented Dec 23, 2015 at 5:04

2 Answers 2


Triangle O'SO fits the description of a 30-60-90 special angled triangle. Therefore, $\angle O'OS = 30^0$

Then, $\angle PYX = 15^0$ [angles at center = 2 times angles at circumference]

  • $\begingroup$ Very nice...........+1 @ Mick $\endgroup$ Commented Dec 23, 2015 at 5:28
  • $\begingroup$ @Ayushakj The solution can be further simplified as follows. Prove that 1) O’Z // XY; 2) $\angle OSO’ = 90^0$; 3) $\triangle PZO’ \sim \triangle PYO$; 4) $OO’ : O’P = OO’ : O’S = 2 : 1$. With the last result and the facts that $OSO’$ is a right angled triangle with $OO’$ being its hypotenuse, you can conclude that $\angle SOO’ = 30^0$ already. In other words, there is no need to use point T. $\endgroup$
    – Mick
    Commented Dec 23, 2015 at 9:13


Notice that $\angle PYX=\angle PYO= \angle OPY$ and $\angle POX=\angle PYO+\angle OPY$, so $$\angle PYX=\frac{1}{2}\angle POX$$ On the other hand, $$\tan \angle POX=\frac{|O'S|}{|SO|}=\frac{a}{a\sqrt{3}}=\frac{1}{\sqrt{3}}\qquad\implies\qquad \angle POX=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}$$


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