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!


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}

| cite | improve this answer | |

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

| cite | improve this answer | |
  • $\begingroup$ How do you get that the angle between $PQ$ and $QR$ is $53^{\circ}$? $\endgroup$ – David Dong Jul 20 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 at 4:02
  • $\begingroup$ @David Dong, I've edited the solution. Let me know if there is still any confusion. $\endgroup$ – SarGe Jul 20 at 5:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.