Solution: McRib
November 18, 2007 pm30 9:02 pm
(This space is for solutions. For the original problem, and for general discussion, click here.)
Given a circle inscribed in a right trapezoid, express the radius of the circle in terms of the bases of the trapezoid.
This space is for answers, and for discussion of answers. If you have a result, look at it. Amazingly simple! Amazingly familiar. Why? I’d really like help understanding.
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JD2718 
Unless I’m missing something in the setup here, I don’t think you can express the radius in terms of the bases.
Explanation: Simplest case is a square with a circle inscribed. Radius is a quarter of the sum of the bases. That is, r = (b1 + b2) / 4. Start from that picture.
Without changing the height of the trapezoid or radius of the circle, change the angle between a base and the non-perpendicular side.
One base gets smaller and the other gets larger. Now imagine what happens as the point of intersection between a base and this angled side approaches the point of tangency between that base and the circle. The length of that base approaches r… and the other base goes to infinity. But r hasn’t changed.
Am I thinking of something different than you’re thinking of?
Huh, threw this into Sketchpad (which is much more fun than what I was doing with it). Having trouble finding a relationship for reasons Ben described above. What am I missing?
But… a circle is determined by three points, so in general you can’t necessarily inscribe a circle in any old right trapezoid. And for those that you can, the diameter of the circle is necessarily the same as the distance between the two parallel sides, so it doesn’t have anything to do with the length of the bases.
I have found that the diameter of the circle is the harmonic mean of the bases, i.e. r = (b1 * b2)/(b1 + b2), but I don’t have a good explanation for this.
Ben, Jackie, Brent…
it can’t be “any old right trapezoid” for the reasons you mention. So, of the right trapezoids that are tangent to a circle at 4 points, express the radius in terms of the bases.
David,
we were thinking to construct the trapezoid, and see if that suggests any reasons. None so far.
Huh. It is true (well, if sketchpad says so it must be true, right?). Now off to figure out why…
I wonder if the reason you’re looking for may be the fact that if you draw two tangent lines from an external point then the segments formed by that point and points of tangency are congruent. In Jackie’s picture, if you call a point of tangency of side AB to the circle P and of side CD to the circle R then BP is congruent to BE (I called their length a) and CE is congruent to CR (and their length is b), so we get a right angle triangle whose sides are 2r, b-a and b+a, and the latter two numbers can be expressed in terms of r, b_1 and b_2, so that after a bit you get David’s solution.
Or was that how you figured it out in the first place? Sorry for two comments. It occurred to me that this may not be an explanation at all.
Wow, interesting. I would not have thought to look at that ratio. How fun! A puzzle!
e,
I did use that fact:
Imagine the diameter perpendicular to the bases.
From the center and the endpoints of the last side of the trapezoid, construct a (right) triangle. Then apply your congruent tangents twice… now we have a right triangle with hypotenuse = unknown side, but divided in pieces of lengths b_1 – r and b_2 – r. The altitude of the triangle (r) is the geometric mean…
e,
My friend “McRib” did exactly what you proposed. I detoured a bit.
So, we’re still looking for a reason? I mean different reason :)
Or a different solution. An explanation for the elegance. Why is the answer so ‘nice’?
I drew some nice-looking trapezoids. I made no progress.
But I think this might be related:
http://www.tamut.edu/~dkern/Presentations/GSketchW/JMM2005.htm
jonathan,
I’m coming to this very late and, with the holiday, this may not be read for awhile, but here are my thoughts on this fascinating problem and wonderful discussion you generated…
First, the result that the radius is the product of the bases divided by their sum is elegant but does not seem that unusual to me (same for the harmonic mean relationship for the diameter). In general, the formula for the the inradius of a polygon is A = rs (obtained by connecting the center to each vertex and adding up the areas of the individual triangles). This formula leads to a general formula for an inradius: r = A/s.
Consider the related problem of inscribing a circle in a right triangle. The formula r = A/s becomes r = (0.5ab)/((a+b+c)/2) = ab/(a+b+c), where r is the inradius, a,b are the legs and c, the hypotenuse.
Ok, so how does this relate to the formula for the right trapezoid? A couple of ways…
First of all, denote the bases of the trapezoid by a and b to avoid subscripts (assume a is less than b). There are then two basic formulas for the area which we will now show are equivalent:
(1) A = h/2(a+b) = r(a+b), since the height of the trapezoid is the diameter of hte incircle.
(2) A = rs, as mentioned above. This again becomes A = r(a+b). This is derived by showing that s = a+b, using the theorem about congruent tangent segments from an external point.
Thus we have, r = A/(a+b). The formula you and others derived is r = ab(a+b). This implies that A = ab, an even more fascinating result in my opinion! This says the AREA OF A RIGHT TRAPEZOID WHICH HAS AN INCIRCLE IS THE PRODUCT OF ITS BASES! Of course, for a triangle the area formula has a factor of one-half.
I derived the formula r = ab/(a+b) similar to e’s method: I dropped an altitude from the end of the upper (shorter) base, producing a right triangle with legs 2r and b-a and hypotenuse (b+a)-2r. Using the Pythagorean Thm and simplifying produces the desired formula.
Finally, why did I compare the right triangle to the right trapezoid? Well, if you extend the non-parallel sides of the right trapezoid, you will obtain a right triangle, one of whose legs is the lower base of your trapezoid. More significantly, THE TRAPEZOID AND THE TRIANGLE SHARE THE SAME INCIRCLE! Someone out there may want to pursue this further…
HAPPY THANKSGIVING EVERYONE!
Some great suggestions for investigation! Thank you Dave.
And I hope your Thanksgiving was happy as well.
I like Dave Marain’s idea of extending the two legs of the trapezoid to make a right triangle with the same incircle.
In my drawing, I drew in the four radii and saw that I got two squares and two kites. The squares are congruent (side length r) and the kites (with sides r and b1 – r in one case, and r and b2 – r in the other case) are SIMILAR!
So r / (b1 – r) = (b2 – r) / r,
and a little algebra gives the formula we want.
Maybe the similarity of the kites is the “why” that you’re looking for?