Geometry puzzle

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Re: Geometry puzzle
Interesting. The answer must be between 6 and 10. Depends on the degree of overlap in the x and y directions and diameters of the circles. Looks like the degree of overlap in the y direction is 2, but I haven't found a way to solve the other three parameters. Will have to chew on it some more later.

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Re: Geometry puzzle
Nope. It's not a square. There's no trick.
Break out your drawing instruments and draw it as accurately as you can  then you can measure x to a decent degree of accuracy.
Those of you who know how can draw it using a CAD program, and so measure x very accurately.
But the best way is to use mathematics or geometry. It's not that difficult.
Break out your drawing instruments and draw it as accurately as you can  then you can measure x to a decent degree of accuracy.
Those of you who know how can draw it using a CAD program, and so measure x very accurately.
But the best way is to use mathematics or geometry. It's not that difficult.

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Re: Geometry puzzle
By eyeball x > 8.
Can we assume all angles are right angles?
I don't understand the significance of the line segment of length 6. Is it given that this line is parallel to the others? I.e., that the vertical lines are all parallel and the horizontal lines are perpendicular to the vertical lines? Can we further assume that the line segment of length 6 passes through the point where the two circles touch?
Can we assume all angles are right angles?
I don't understand the significance of the line segment of length 6. Is it given that this line is parallel to the others? I.e., that the vertical lines are all parallel and the horizontal lines are perpendicular to the vertical lines? Can we further assume that the line segment of length 6 passes through the point where the two circles touch?

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Re: Geometry puzzle
If you square a circle then add another circle, and square them both, its always a square

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Re: Geometry puzzle
Yes x > 8.Anaxagoras wrote: ↑Wed Sep 16, 2020 10:30 am By eyeball x > 8.
Can we assume all angles are right angles?
I don't understand the significance of the line segment of length 6. Is it given that this line is parallel to the others? I.e., that the vertical lines are all parallel and the horizontal lines are perpendicular to the vertical lines? Can we further assume that the line segment of length 6 passes through the point where the two circles touch?
The outer rectangle is a true rectangle (rightangled corners and straight sides) with short side length 8.
The two circles are true circles. They touch each other tangentially (just touch without overlapping). One circle tangentially touches one short side and one long side of the rectangle, the other circle tangentially touches the remaining two sides.
The line shown with length 6 is parallel to the short sides of the rectangle. The line's ends are points on the circles, and the line passes through the point where the circles touch.

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Re: Geometry puzzle
Yeah, there may be some principle of geometry that allows one to solve for x, but I don't know what it is.

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Re: Geometry puzzle
Most people would agree that the Pythagorean theorem is a fundamental part of geometry. As Abdul suggested, the trick with most of these sorts of puzzle is to draw a few lines and then apply Pythagoras. The real trick is to know which lines to draw!Anaxagoras wrote: ↑Wed Sep 16, 2020 3:30 pm Yeah, there may be some principle of geometry that allows one to solve for x, but I don't know what it is.

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Re: Geometry puzzle
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.

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Re: Geometry puzzle
Without looking for the answer I'm going to guess that it's 9, but I can't show my work.
Either that or it's not a whole number.
Either that or it's not a whole number.

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Re: Geometry puzzle
Abdul Alhazred wrote: ↑Thu Sep 17, 2020 3:13 pmIt's our old buddy correlation with population size at work.
More high school geometry students available? :wink:

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Re: Geometry puzzle
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.)

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Re: Geometry puzzle
I’m an idiot
Last edited by robinson on Sat Sep 19, 2020 6:05 am, edited 1 time in total.

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

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Re: Geometry puzzle
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.
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.
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.

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Re: Geometry puzzle
Finally got it. Messy and convoluted, but it works. That was fun.
https://i.imgur.com/60UQlA6.jpg
https://i.imgur.com/60UQlA6.jpg

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