Sure you can. For instance, if you set A=3/4 and B=5 then you can isometrically embed the figure above into a surface that looks something like this (albeit a tiny bit smoother): [embedding]

In other words, since this problem seems to be about challenging your assumptions, I'll ask you to challenge your own... ;) New question: does there exist an assignment for A and B such that the figure above can be embedded isometrically in a space of constant curvature?

I'm not trying to be too sneaky (hence the title of the post). You can arrive at A=9 and B=5 using the Pythagorean theorem. However, the rule of similar triangles forces A=3/4, hence a contradiction. A variant of this blunder was found on some SAT exam.

I'm not sure if there exists an assignment of A and B which allows such an embedding. I'll think about it when I have more spare time on my hands =)

Suppose we draw a triangle on a sphere. Then it's possible to draw one with three 90 degree angles (one vertex on the north pole and two vertices on the equator somewhere). Keenan's question is whether there exists values A and B and such a surface (not necessarily a sphere) such that the figure makes sense.

Consider a sphere with radius 1/pi. The great circumference of the sphere has length 2. More importantly, the distance between the north and south poles is 1.

Let the vertex connecting the segments of lengths 3 and 4 be located at the north pole.

The segment of length 4 will loop around the prime meridian (and 180th meridian) twice and end up back at the north pole.

The segment of length 3 will loop around at a different longitude 1.5 times and end up at the south pole.

The segment of length B connects the north and south poles. Thus we see that B can be any positive odd integer.

We also know that the segment of length 8 starts at the north pole and follows due south. Since it has an even integer value, it must loop around some great circle arc and end up back at the north pole.

Using the same reasoning, the segment of length 5 must connect the north and south poles.

Thus we can conclude that A must be a positive odd integer since that segment also connects the north and south poles.

Yisong, I followed you till the very end. You said that the end of 8 near 5 was the north pole, therefore the end of 5 near A was the south pole. We already determined that the end of 3 near A was the south pole. This means that A has to be even, right?

## 10 comments:

A=9 and

B=5

... but these triangles aren't physically possible to construct, are they?

Sure you can. For instance, if you set A=3/4 and B=5 then you can isometrically embed the figure above into a surface that looks something like this (albeit a tiny bit smoother): [embedding]

In other words, since this problem seems to be about challenging your assumptions, I'll ask you to challenge your own... ;) New question: does there exist an assignment for A and B such that the figure above can be embedded isometrically in a space of

constantcurvature?-Keenan

(Actually, that figure has a glaring error, but you get my point.)

-Keenan

Yah what lance said, then anonymous here confused the heck out of me. I think I don't even get what I'm not getting here.

Whats wrong with A=9 B=5?

I'm not trying to be too sneaky (hence the title of the post). You can arrive at A=9 and B=5 using the Pythagorean theorem. However, the rule of similar triangles forces A=3/4, hence a contradiction. A variant of this blunder was found on some SAT exam.

I'm not sure if there exists an assignment of A and B which allows such an embedding. I'll think about it when I have more spare time on my hands =)

Suppose we draw a triangle on a sphere. Then it's possible to draw one with three 90 degree angles (one vertex on the north pole and two vertices on the equator somewhere). Keenan's question is whether there exists values A and B and such a surface (not necessarily a sphere) such that the figure makes sense.

Here's a pretty degenerate solution.

Consider a sphere with radius 1/pi. The great circumference of the sphere has length 2. More importantly, the distance between the north and south poles is 1.

Let the vertex connecting the segments of lengths 3 and 4 be located at the north pole.

The segment of length 4 will loop around the prime meridian (and 180th meridian) twice and end up back at the north pole.

The segment of length 3 will loop around at a different longitude 1.5 times and end up at the south pole.

The segment of length B connects the north and south poles. Thus we see that B can be any positive odd integer.

We also know that the segment of length 8 starts at the north pole and follows due south. Since it has an even integer value, it must loop around some great circle arc and end up back at the north pole.

Using the same reasoning, the segment of length 5 must connect the north and south poles.

Thus we can conclude that A must be a positive odd integer since that segment also connects the north and south poles.

Yisong, I followed you till the very end. You said that the end of 8 near 5 was the north pole, therefore the end of 5 near A was the south pole. We already determined that the end of 3 near A was the south pole. This means that A has to be even, right?

I think the image from the URL below will clear things up. Blue indicates north pole, and red indicates south pole.

http://www.yisongyue.com/blogger/stupid_math_poles.png

Yisong, that solution was

extremelydegenerate. Shame on you!=)

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