Segment PQ is a chord common to two circles and it bisects
Angle(RPT), where R
and T lie on the circles, as shown. Each of the chords P R
and P T is cut by the other circle at points S
and U. Prove that R S = T U.
A basic geometry problem involving circles
circleseuclidean-geometrygeometry
Related Solutions
This is true by symmetry. The figure is symmetric with respect to reflection in the line joing the centres; if the line joining the two intersection points weren't perpendicular to that line, it would break the symmetry.
A symmetry argument shows that the top two blue circles are tangent not just to the green circles on the same side of the chords, but also to the top green circle. This is because a reflection about a diameter that passes through, say, the center of the top right blue circle, must map that blue circle to itself, and the rightmost green circle to the topmost green circle, and vice versa.
Consequently, there is a horizontal tangent line that can be drawn so that the top five circles forms a third sector with all the corresponding tangencies preserved. This also implies that a circle centered at the origin that is externally tangent to all three green circles will have the same radius as the blue circle, since a reflection about a line joining the centers of the top green circle with the rightmost green circle will map the center of the top right blue circle to the origin.
Next, it is easy to determine the radius of the green cirles. Assuming without loss of generality that the large circle has unit radius. Then the radius of the green circle must satisfy
$$\frac{g}{2-g} = 1 - 2g, \implies g = \frac{3 - \sqrt{5}}{2}. \tag{1}$$
This in turn means that the blue circle has radius $$b = 1 - 2g = \sqrt{5} - 2. \tag{2}$$
The radius of the red circle $r$ is the most difficult to determine. One way is to place the figure on the coordinate plane and compute the coordinate of the center of the top right blue circle, which is $$(x_b, y_b) = \left(\sqrt{10 \sqrt{5} - 22}, 2(\sqrt{5} - 2) \right), \tag{3}$$ the proof of which I leave as an exercise (hint: compute the coordinate of the intersection of the horizontal chord with the chord on the right). Then $r$ must satisfy
$$\begin{align} x_r^2 &= (1-r)^2 - (b+r)^2, \\ (x_r - x_b)^2 + (b - r)^2 &= (b + r)^2, \end{align} \tag{3}$$ where $x_r$ is a nuisance variable representing the $x$-coordinate of the top right red circle. This gives us the unique nontrivial solution $$x_r = 4 \sqrt{\frac{\sqrt{5}-2}{5}}, \quad r = \frac{3 - \sqrt{5}}{10}. \tag{4}$$ hence $g/r = 5$.
For fun, I also made a little animation.
Best Answer
Notice that, $\angle RQU=180-\angle RPU=180-\angle SPT=\angle SQT$ and thereafter $\angle RQS=\angle UQT$.
In triangle $\triangle QRS$ and $\triangle QUT$, $QR=QU$, $QS=QT$ and $\angle RQS=\angle UQT$.
Hence, $\triangle QRS\cong \triangle QUT$ by $S--A-S$ criteria of congruence and thereafter $RS=TU$.