Two circles, both with the same radius $r$, are placed in the plane without intersecting each other. A line in the plane intersects the first circle at the points $A,B$ and the other at points $C,D$, so that $|AB|=|BC|=|CD|=14\text{cm}$. Another line intersects the circles at $E,F$, respectively $G,H$ so that $|EF|=|FG|=|GH|=6\text{cm}$. Find the radius $r$.
Problem
Source: Baltic Way 1993
Tags: geometry unsolved, geometry
subham1729
15.06.2012 17:46
First, note that the centres O1 and O2 of the two circles lie on different sides of the line EH — otherwise we have r < 12 and AB cannot be equal to 14. Let P be the intersection point of EH and O1O2 . Points A and D lie on the same side of the line O1O2 (otherwise the three lines AD, EH and O1O2 would intersect in P and |AB| = |BC| = |CD|, |EF| = |FG| = |GH| would imply |BC| = |FG|, a contradiction). It is easy to see that |O1O2| = 2 · |O1P| = |AC| = 28 cm. Let h = |O1T | be the height of triangle O1EP. Then we have h2 = 142 − 62 = 160 from triangle O1TP and r2 = h2 + 32 = 169 from triangle O1TF. Thus r = 13 cm.
sunken rock
16.06.2012 18:15
As before, $EH$ separates $O_1, O_2$, the centers of the two triangles. Let $M, N, P$ be midpoints of $O_1O_2, O_1M, EF$ respectively. Obviously, $F$ is midpoint of $MP$ and $N$ the projection of $B$ onto $O_1O_2$. Consequently, in the right-angled triangle $\triangle MNF$: $MN=7, MF=3,\implies NF=\sqrt{40}$. As $NF$ is the midline of $\triangle PO_1M$, $PO_1=2\cdot\sqrt{40}$. In the right-angled triangle $\triangle PO_1F: PF=3, PO_1=2\cdot\sqrt{40}$, hence $O_1F^2=9+160\iff r=13$, done. Best regards, sunken rock