Proof: circumference of circle inscribed in a triangle < perimeter of the triangle

[fleazo]

Homework Statement

A circle is inscribed in a triangleHere is a picture Picture of circle inscribed in triangle, not necessarily to scaleWhich is larger: the circumference of the circle, or the perimeter of the triangle?

Homework Equations

C=∏D (D=diameter of the circle, C=circumference of circle)
P=AB+BC+CA (P stands for perimeter, refer to the linked photo to see A,B, and C)

The Attempt at a Solution

This is taken out of Nova's Math GRE Bible. Their solution is as follows, but I have a dispute with the solution:

"From the figure, it is clear that to go from one point on the circle, say, point P to another point, say, point
Q, the shortest available path is the arc PQ. Hence, arc PQ < PA + AQ. Similarly, arc QR < QB + BR, and
arc RP < RC + CP. Summing the three inequalities yields arc PQ + arc QR + arc RP < (PA + AQ) +
(QB + BR) + (RC + CP). The right side of the inequality is the perimeter of the triangle ABC, and the left side is the circumference of the circle. Hence,
Column A is greater than Column B, and the answer is [the perimeter of the triangle is larger]."What I'm disputing is their statement: "it is clear that to go from one point of the circle P to Q the shortest available path is the arc PQ" Wouldn't the shortest path be the chord PQ (because the shortest path between two points is a straight line, which, in that case would be the chord PQ NOT the arc)? I know that, for example the chord PQ < PA + AQ because we can form a triangle APQ and we know each side has to be less than the sum of the lengths of the other side. We can continue forming these small triangles and then end up with, just as they were doingPQ + QR + RP < PA + AQ + QB + BR + RC + CP

where the right side is the perimeter of the big triangle, but now the left side is only the perimeter of a smaller triangle, PQR, that is inscribed in the circle!Where do you get the relationship that the arc(PQ) < PQ + AQ as they are stating in this proof?Thank you! I am very confused by their proof

 

[lewando]

Yes, the chord would be the shortest path--they should have made it clear that they were only comparing the arc path to the triangle path.

Looks like they are not proving arc(PQ) < PQ + AQ. If you want to prove this, then you could:

1. first prove that the triangle is a equilateral triangle.
2. demonstrate that the 3 arcs are equal.
2a. [edit-- and the circle bisects each side of the triangle at the point of contact].
3. relate the length of one of those arcs (1/3 of the circle circumference) to the length of a side of the triangle.

I can help with 3.

 

[fleazo]

Thanks for the reply. Another confusion: about the triangle being equilateral - why would we assume this? You can inscribe a circle in any sort of triangle right?

 

[lewando]

Yes you are right. I was only looking at the equilateral case. You want a more general proof.

 

[alan2]

Correct, it need not be equilateral.

 

[fleazo]

THanks for the clarification. To be honest, I am prepping for the GRE, and I'm just trying to look at that problem and think, how would I see the general proof really quickly (since I can't make any assumptions like it being equilateral or anything). This was listed as an easy problem so my confidence is just really shot and I feel pretty dumb that I don't see the solution, and more than that that I don't understand exactly how NOVA is proving it in their given solution :(

 

[alan2]

Some advice. I've taught a lot of test prep. The GRE will never ask you to prove anything like this, particularly since it is a multiple choice test. They are more interested in your ability to quickly spot relationships. In this case, you have 3 points and you are comparing paths which connect these three points. Clearly, the shortest path is that which connects the 3 points with line segments, i.e. the triangle inscribed in the circle (which is not shown). Since the circumference lies entirely between this shortest path and the path defined by the perimeter of the large given triangle then the circumference of the circle is smaller than the perimeter of the triangle in question.

My advice, take it or leave it, is to avoid prep books published by others. Get the official book written by the test maker which includes actual retired test questions. Do every problem in that book, trying not to actually write anything down. You will start to see patterns in the questions and should be able to answer most of them with a bit of critical reasoning. This is not a math test! The same advice goes for the verbal sections.

 

[fleazo]

alan2 - thank you so much for that explanation, I completely see the answer now and I feel really foolish for not seeing it before


The problem with this question was it was one of the quantitative comparison questions where they give the figure and say , which is larger?

a) circumference of circle
b) perimeter of triangle
c) they are equal
d) can not be determined


So because of the "can not be determined" I feel I need a proof to be sure about any answer I write down. Like on this one, I "felt" that the perimeter is larger, but I felt, how can I prove this to myself before I select that answer. Again, man do I feel dumb for not seeing it now, and thanks so much for pointing out the reasoning


About the test prep books... yeah I grabbed the one ETS produced and I do like it, I feel like I can be most confident with it because well, they make the exam. But I was turning to the princeton review just because they have so many questions (100+ just for the geometry section)

 

[lewando]

For what its worth-- this is as close as I can find for a general proof:

This is a general arc-corner scenario. By symmetry, we need to only consider s and d--need to show that s < d.

The formula for arclength is s = rθ (θ in radians).

Applying that formula, need to show that rθ < d:

Re-organizing: θ < d/r

Note: tan(θ) = d/r

If θ < tan(θ), then proven (which it is over the interval 0 < θ < pi/2 ).