Too Ugly?

Geometry Level 1

In a large circle of diameter 4, we draw two circles that are each internally tangent at A A and B . B. We then draw the common interior tangent C D . CD. If A C AC intersects the large circle at E E and B D BD intersects the large circle at F , F, what is the distance E F ? EF?

3 3 π \pi 2 3 2 \sqrt{3} 4 4 Depends on the length of A B AB

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Calvin Lin Staff
Oct 28, 2016

We will repeatedly use the following lemma in homothety .

Lemma: Given 2 circles that are internally tangent at T T , if we draw any line through T T cutting the circles at A A and B B respectively, then the tangent at A A is parallel to the tangent at B B .

Proof: Consider the expansion mapping about point T T that brings the smaller circle to the larger circle.
Then, we see that map will bring point A to point B. Also, by considering the tangent to the circle, we see that it will bring the tangent at A to the tangent at B.
Since the expansion mapping keeps the slope of a line constant, this means that the tangent at A will have the same slope as the tangent at B, thus they are parallel. _\square

Now, on to the original problem.
Construct the tangents at E and F.

Step 1: From the above lemma, the tangent at E is parallel to the tangent at C.
Step 2: From the above lemma, the tangent at F is parallel to the tangent at D.
Step 3: From the previous steps, the tangent at E is parallel to the tangent at F.
Step 4: EF is a diameter of the circle.

Note: As observed by Michael, EF is perpendicular to CD. This follows because the diameter EF is perpendicular to the tangent at E, which is parallel to CD.

Homothety is a nice subject in geometry---I see the proposed wiki that should be completed

Michael Mendrin - 4 years, 7 months ago

Log in to reply

It's my favourite Euclidean Geometry technique. (I even talk about it in my interviews!)

I would love to help develop out the page. Here's my proposed outline

  1. Explanation of general concept
  2. Basic usages - Ratios, Tangency, Collinearity
  3. Applications in circles (esp collinearity, concurrency)
  4. Applications in similar triangles (median triangle).

Calvin Lin Staff - 4 years, 7 months ago

Log in to reply

I know I haven't been very helpful about creating wiki pages even though I keep talking about doing just that. Maybe I'll make this my first effort. Well, actually, second? I think I have the time now, so let me review the literature.

Michael Mendrin - 4 years, 7 months ago

Okay I added a few lines

Wen Z - 4 years, 7 months ago

Log in to reply

You mean the introductory block of that wiki?

Michael Mendrin - 4 years, 7 months ago

That's great!

We are using this doc to plan out the wiki, and will start writing it up over the next week :)

I would love to hear your comments / feedback.

Calvin Lin Staff - 4 years, 7 months ago

I've just looked up homothety. God it's so simple. Lovely solution.

Desmond Campbell - 4 years, 7 months ago

Log in to reply

Yes! It's a simple concept which allows us to gain a lot of insight whenever a problem has a bunch of similarity to consider.

Are there any other applications that come to mind?

Calvin Lin Staff - 4 years, 7 months ago

The document you provided is absolutely BRILLIANT!!! I found it really instructive and interesting. To add to the amazement, this very question came in our this year's RMO examination.

Jitarani Nayak - 3 years, 7 months ago
Michael Mendrin
Oct 27, 2016

Just for reasons of symmetry alone, the distance can't be anything else but the diameter of the circle.

Oh come on ....

How is that implied? It could be "depends on AB".

Calvin Lin Staff - 4 years, 7 months ago

I'd like to see the reasoning.

Desmond Campbell - 4 years, 7 months ago

Log in to reply

Since it's implied that the length EF is invariant regardless of where the two circles are drawn, I can choose two circles that are exactly half the size of the large circle. Then the tangent as well as EF will be perpendicular to each other, meeting at the center of the circle.

In fact, the tangent as well as EF will always be perpendicular to each other, while the intersection point will depend on the two circles chosen.

Michael Mendrin - 4 years, 7 months ago

At first I didn't see what you meant when you said 'implied invariant' but I get it now, you mean implied by the answers. However as the other commenter says, that isn't the case.

Desmond Campbell - 4 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...