Calvin and Hobbes by Bill Watterson for April 27, 1987

April 26, 1987

April 28, 1987

Transcript:

Book: Point A is twice as far from point C as point B is from A. If the distance from point B to point C is 5 inches, how far is point A from point C? Calvin: The living dead don't need to solve word problems.

Hardcore math problem for a 6-year-old. It depends on the angle BAC.Here is a graph of the set of the solutions with the value of AC depending on the value of the angle BAC in radians : http://i.imgur.com/PhoSi.pngI’d rather walk dead than having to solve this by hand, searching for limits and tangents.

If they are in a straight line, it should be 10 inches, or 3 1/3 inches. I believe the possible answers lie on a hyperbola, so the answer may be anything over 3 1/3 inches. A diagram, or more constraints, sure would help.

The lengths of the sides are X, 2X, and 5. We are asked to find 2X. The term “distance” is used, so the triangle inequality applies. Therefore:

X + 2X >= 5

X + 5 >= 2X

5 + 2X >= X

Reducing, and solving for 2X:

2X >= 3 1/3

2X <= 10

2X >= -10

So all we know, without further assumptions is that AC is between 3 1/3 and 10, inclusive.

Note that this analysis does not assume and particular distance function. In particular it does not assume a distance function derived from a euclidian norm. But I doubt Calvin’s teacher could even tell you what that means.

## DragonWizzard almost 9 years ago

This is how I fell when I get these math problems.

## ilovemypapillon over 8 years ago

I don’t think that is possible to solve whether you are dead or alive

## comicmath about 8 years ago

Maybe with a diagram

## overlived over 7 years ago

two inches less than a foot and i’m in zombie town

## jeanmichel over 7 years ago

Hardcore math problem for a 6-year-old. It depends on the angle BAC.Here is a graph of the set of the solutions with the value of AC depending on the value of the angle BAC in radians : http://i.imgur.com/PhoSi.pngI’d rather walk dead than having to solve this by hand, searching for limits and tangents.

## WilliamBill over 7 years ago

Isn’t it a straight line? 10"?

## bmonk over 7 years ago

If they are in a straight line, it should be 10 inches, or 3 1/3 inches. I believe the possible answers lie on a hyperbola, so the answer may be anything over 3 1/3 inches. A diagram, or more constraints, sure would help.

## yow4zip Premium Member over 7 years ago

That’s why zombies eat brains.

## pink54836 over 5 years ago

um i think this just goes in circles

## VRAssassinCreed almost 5 years ago

10

## the calvinosaurus that calvin wanted to discover over 4 years ago

yeah 10

## BarrelO'Molasses Premium Member almost 3 years ago

10"

## zahaanr7 about 2 years ago

10"

## Comics Man almost 2 years ago

10"

## donerito almost 2 years ago

It’s only 10" if you make assumptions (and we all know that makes an ass-of-u-&-me).

## MarkFaxas about 1 year ago

AC = 2ABBC = 5BC = ABAB = 5AC = 2AB = 2 × 5 = 10AC = 1010 inches.

## divad27182 11 months ago

The lengths of the sides are X, 2X, and 5. We are asked to find 2X. The term “distance” is used, so the triangle inequality applies. Therefore:

X + 2X >= 5

X + 5 >= 2X

5 + 2X >= X

Reducing, and solving for 2X:

2X >= 3 1/3

2X <= 10

2X >= -10

So all we know, without further assumptions is that AC is between 3 1/3 and 10, inclusive.

Note that this analysis does not assume and particular distance function. In particular it does not assume a distance function derived from a euclidian norm. But I doubt Calvin’s teacher could even tell you what that means.

## Comic Lover 2020 Premium Member 11 months ago

AB: 5"

BC: 5"

AC: 10"