Geometry puzzle

[Abdul Alhazred]

Interesting that ceptimus posted this over at ISF this morning and had two correct answers in less than an hour. I'll draw no conclusions.

It's our old buddy correlation with population size at work.

[DJ]

Interesting that ceptimus posted this over at ISF this morning and had two correct answers in less than an hour. I'll draw no conclusions.

It's our old buddy correlation with population size at work.

More high school geometry students available?

[robinson]

Math is hard

[robinson]

Geometry even harder

[Witness]

A back of the envelope calculation gives me 9. If that's correct I'll write up my solution. (If not I'll just hang my head in shame.)

[robinson]

I’m an idiot

[ceptimus]

A back of the envelope calculation gives me 9. If that's correct I'll write up my solution. (If not I'll just hang my head in shame.)

Yes, 9 is the correct answer.

[Abdul Alhazred]

A back of the envelope calculation gives me 9. If that's correct I'll write up my solution. (If not I'll just hang my head in shame.)

Yes, 9 is the correct answer.

One of you please explain how to get there.

[Witness]

One of you please explain how to get there.

Here you go:

[robinson]

Jesus fucking christ almighty

[robinson]

It's like posting the Riemann Hypothesis in a puzzle forum. Then laughing like a maniac later that night.

[ceptimus]

Here's a simple geometric proof by Pan Narrans (a poster at the Freethought Forum). Pan was actually the first poster to solve my posting of the puzzle, but I posted the puzzle there about the same time as here, and before I did at ISF, so the ISF guys were the fastest. I didn't invent the puzzle by the way - I first saw it posed in a YouTube "recommended for you" still - but I didn't watch the video and solved it myself instead.

geometryPuzzle_3.gif

AD = AE = AG = AK (the radius of the small circle)

similarly

BD = BF = BH = BL (the radius of the large circle)

Triangle ADK is an isosceles triangle, with A as its apex. The same goes for triangle BDL, with apex B. The "bases" of the two isosceles triangles (the non equal sides) are KD and DL. Note that the line AC is the same length as half the base of the small triangle plus half the base of the large triangle

AC = (KD + DL)/2, and we know that KD + DL = 6, so AC = 3

Now AG + AC + BH = 8 (the height of the rectangle)

Subtracting AC = 3 gives AG + BH = 5 (the sum of the radii)

Since AD + BD = AG + BH = 5, we have a 3/4/5 right triangle ABC with a hypotenuse (AB) 5 and one side (AC) 3, the other side (BC) must be 4.

X = AE + BF + BC = AG + BH + BC = 5 + 4 = 9

Pan did a clever proof using trigonometry and simultaneous equations, similar to Witness's, first, but then followed it up with this simpler geometric one.

The way I did it was something of a cheat. I reasoned that the puzzle wouldn't have been set unless it had a neat constant answer - not some horrible formula depending on the sizes of the circles drawn. So I figured I could make the smaller circle as small as I wanted without changing the answer. If you draw it with the small circle diameter zero, then you just have a rectangle with short side 8, and the large circle touching the bottom and right hand side of the rectangle and passing through the top left hand corner. The left hand edge of rectangle cuts a chord length 6 across the circle. Then the 3.4.5 triangle is immediately obvious, and you get the solution in a flash.

[Bruce]

Finally got it. Messy and convoluted, but it works. That was fun.

[Bruce]

It was quite satisfying to see a 3,4,5 right triangle emerge from my mess.

[Abdul Alhazred]

It was quite satisfying to see a 3,4,5 right triangle emerge from my mess.

[shemp]

It was quite satisfying to see a 3,4,5 right triangle emerge from my mess.

Yeah, sometimes they emerge from my ass when I'm dropping a log.