# A circle rolling along the base of an isosceles triangle has constant arc length cut out by the lateral sides.

The problem is from Kiselev's Geometry Exercise 582:

A circle of the radius congruent to the altitude of a given isosceles triangle is rolling along the base. Show that the arc length cut out on the circle by the lateral sides of the triangle remains constant.

[Edited] The problem is very vague, but the correct version of it is that the circle should either pass through the top vertex or both lateral sides.

My attempt was to draw a line parallel to the base and passing through the top vertex. Then the case when the circle passes through the top vertex is easy since the side angle formed by the intersection of the circle and a lateral side is the same as the side angle of the given isosceles triangle. However, I could not derive the same conclusion when the circle intersects both lateral sides.

Any help would be greatly appreciated.

• What is the angle when the circle is near the center of the triangle, but not exactly over it? Is is the same as in the center? Maybe the writer assumed you would make the height larger than the base, and the identity only applies when both lateral sides touch the circle. Jul 29, 2020 at 0:27
• I just verified that my suggestion was correct. If the center of the circle stays within one radius from the center of the triangle, the identity holds. The writers of the book must not have thought about triangles as wide as the one you use in your example. Jul 29, 2020 at 0:39
• A requirement seem to be that both sides intersect with the circle - so the arc has always an angle $\beta$
– Moti
Jul 29, 2020 at 0:49
• Thank you everyone; I have also checked it using Geogebra. Now I still could not solve the problem. I would really appreciate if one could provide a solution (I will provide my attempt as well). If it is more appropriate for the question to be asked separately, I will do so accordingly. Jul 29, 2020 at 1:02
• I think you should keep this question, but remove the part about possibly misunderstanding the question, and shift the focus to how to solve it. I personally have no clue how to do this without analytic geometry, but someone else will. Jul 29, 2020 at 2:03

Given an isosceles triangle $$\triangle ABC$$ with apex $$B,$$ extend sides $$AB$$ and $$CB$$ to form a congruent mirror-image triangle $$\triangle DBE$$. The circle then is inscribed between the parallel bases of the triangles. Let the intersections of the legs of both triangles with the circle be $$P,Q,R,S$$ as shown in the figure.
Provided that the apex $$B$$ is inside the circle, a theorem about the arcs intercepted by two intersecting chords of a circle says that the sum of the angle measures of arcs $$\stackrel{\frown}{PQ}$$ and $$\stackrel{\frown}{RS}$$ is $$2\times \angle CBE.$$
Note that the arcs $$\stackrel{\frown}{QR}$$ and $$\stackrel{\frown}{PS}$$ are congruent.
• I was looking for something that would stay constant as we rolled the circle to one side or the other. I did not know any facts about a single angle (such as $\angle ABC$) at an arbitrary point inside a circle, but I recalled the fact about the intersecting chords. I did not actually need the other triangle; I could have used a parallel line through $B,$ but the symmetry of the second triangle looked interesting and turned out to be useful. Sometimes it's just about thinking about a lot of different ways to look at it until something works. Jul 29, 2020 at 3:23