# Infinite sequences in geometry

My math teacher gave these 2 bonus questions for extra credit. I have no idea where to even begin - could I get a few pointers to lead me in the right direction?

1. Each circle in an infinite sequence with decreasing radii is tangent externally to the one following it and to both sides of a given right angle. What is the ratio of the area of the first circle to the sum of the areas of the other circles in the sequence?

-Obviously the ratio of the radius of the first circle to the sum of the radii of the other circles is $1:\sqrt{2} -1$, but i am unsure of how to find the ratio for the areas.

2. Consider an isosceles triangle ABC with two sides equal to 1 and $\angle A, \angle C=72^\circ$. Bisect $\angle A$ with bisector AD and call its length $I_{1}$. Bisect $\angle D$ with bisector DE and call its length $I_{2}$. Continue this process and determine $I_{1} + I_{2} + I_{3} + ...$

-All these triangles are similar, but I'm not sure of the ratio in which ABC is greater than ADC due to the fact we haven't learned trigonometry yet, so how should I go about solving this problem?

• I'm trying to visualize the process you're describing in 2, and I can see it a couple of different ways. Can you describe again, or maybe supply a a diagram? – bob.sacamento Mar 3 '14 at 2:27
• Here's a picture of question 2: i.imgur.com/nS3vsT1.png – Andrew Nguyen Mar 3 '14 at 2:33
• Is it possible that those angles should be 72 instead of 75? With that change, the problem becomes much easier. – bob.sacamento Mar 3 '14 at 5:27
• Ah I'm sorry - you're correct. I've edited my original question. – Andrew Nguyen Mar 3 '14 at 5:28
• OK, with that change in the angle, you should be able to see that each time you do a new iteration, the new length is AC/AB times the previous length. It's another sum of a power series. – bob.sacamento Mar 3 '14 at 18:31

## 1 Answer

1: Let's say your largest circle has radius one. Then from the corner to the center of the circle is a distance of $\sqrt2$. And the distance from the corner to the rim of the circle is $\sqrt 2 -1$. Now, let's say that the second circle has radius (that's radius, not diameter) $a$. I will leave it to you to work out that the third circle has radius $a^2$, the fourth has radius $a^3$ etc. Then the distance from the corner to the rim is

$2\sum_{n=1} a^n = \sqrt 2 -1$

You can get $a$, and any ratio from there. The ratio of areas is related to the square of the ratio of radii.

• I only know the ratio of the radius of the large circle to the SUM of the radii of all the other circles in the sequence, by using a $1-1-\sqrt{2}$ triangle. I don't know how I would find out the ratio of any 2 tangent circles in this sequence. – Andrew Nguyen Mar 3 '14 at 2:35
• Ah, I see. Answer edited. – bob.sacamento Mar 3 '14 at 3:14
• Thanks, I got the answer! For those in my class who might stumble upon this in the future -- examine the line $y = x$, let $A$ be the first point where this line intersects the large circle, and let $B$ be the second point where it intersects. These two lines are proportional with the same ratio as the radii of the first circle and the next. Once you find $OA$ and $OB$, the ratio $r_{1} : r_{2} = OB : OA$ can be used to find the radius of the smaller circle. Combined with what bob.sacamento said, this should be enough to figure it out :) – Andrew Nguyen Mar 3 '14 at 5:17