From HMMT:

Triangle $\triangle PQR$, with $PQ=PR=5$ and $QR=6$, is inscribed in circle $\omega$. Compute the radius of the circle with center on $QR$ which is tangent to both $\omega$ and $PQ$.

I haven't made much progress. I've set $QS=x$ and $SR=y$ to try for Law of Cosines, since we know $\cos\angle QSR$, but that really hasn't lead anywhere. With Ptolemy's, I 've found that $PS=\displaystyle\frac{5(x+y)}{6}$, but unfortunately $PS$ isn't colinear with anything useful (like the line connecting the centers). I also haven't really been able to use the tangent properties.

Hints beyond what I've done or any useful insights would be greatly appreciated!


2 Answers 2


enter image description here

Given that $|PQ|=|PR|=5,\ |QR|=6$, the area, the height and the circumradius of $\triangle PQR$ are $S=12$, $|PF|=4$ and $R_0=\tfrac{25}8$, respectively. Let $\angle PQR=\alpha$, $\angle FOE=\phi$.

Assuming that the center of the circle $O_t\in QR$, we must have $|DQ_t|=|EQ_t|=r$.

\begin{align} \sin\alpha&=\frac{|PF|}{|PQ|} =\frac45 ,\\ |OF|&=|PF|-R_0=\tfrac78 \tag{1}\label{1} . \end{align}

We have two conditions for $r$, $\phi$:

\begin{align} |QO_t|+|FO_t|&=\tfrac12\,|QR| \tag{2}\label{2} ,\\ \frac r{\sin\alpha} + (R_0-r)\sin\phi &=3 \tag{3}\label{3} ,\\ (R_0-r)\cos\phi&=|OF| \tag{4}\label{4} . \end{align}

Excluding $\phi$ from \eqref{3},\eqref{4} and using known values, we get

\begin{align} r&=\frac{20}9 . \end{align}


I've done this using coordinate geometry. Let the sides of the triangle be along the sides as shown in the figure. enter image description here The angle between $PQ$ and $QR$ is $53°$ (approximately) because of the $(3,4,5)$ triangle formed by the sides $PQ,QB,BP$ where $B$ is midpoint of $QR$. So, the equation line along $PQ$ is $3y=4x$.

Now, let the centre and radius of the unknown circle be $C(h, 0) $ and $r$ respectively. We get two constraints from these,

  1. Perpendicular distance from $C$ to the line $PQ$ is $r$.

  2. $\displaystyle AC=\frac{25}{8}-r$.

You've two equations and two variable and may proceed now

  • $\begingroup$ How do you get that the angle between $PQ$ and $QR$ is $53^{\circ}$? $\endgroup$
    – mpnm
    Jul 20, 2020 at 3:09
  • 1
    $\begingroup$ The $(3,4,5)$ Pythagorean triplet has well know trigonometric inverses(they occur a lot in elementary Physics exercises); however, explicitly assuming the angle does not seem fair and I think @SarGe should offer an explanation. $\endgroup$
    – Manan
    Jul 20, 2020 at 4:02
  • $\begingroup$ @David Dong, I've edited the solution. Let me know if there is still any confusion. $\endgroup$
    – SarGe
    Jul 20, 2020 at 5:20

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