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# Mathematics Brain Teasers.

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Thanks for that, Anton. I forgot about the redundant subtraction of the zero should have inserted back into my solution.

Here is my next problem based on geometry for a change.

Problem: A rectangle whose long side is 2.4 times the length of its short side is inscribed in a circle of diameter 130m (see below). What is the area inside the circle but outside the rectangle?

Edited by Ozymandias
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49 minutes ago, Ozymandias said:

Thanks for that, Anton. I forgot about the redundant subtraction of the zero should have inserted back into my solution.

Here is my next problem based on geometry for a change.

Problem: A rectangle whose long side is 2.4 times the length of its short side is inscribed in a circle of diameter 130m (see below). What is the area inside the circle but outside the rectangle?

(zero was originally missing in the official answer and I couldn't stand zero got skipped that way. So I added that *important* substraction. Whenever we're asked to count to 10, we're supposed to start from zero, right? ^^ I cherish zero. I even regularly list things from the zeroth: 0th, 1st, 2nd, etc)

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1 hour ago, Ozymandias said:

Thanks for that, Anton. I forgot about the redundant subtraction of the zero should have inserted back into my solution.

Here is my next problem based on geometry for a change.

Problem: A rectangle whose long side is 2.4 times the length of its short side is inscribed in a circle of diameter 130m (see below). What is the area inside the circle but outside the rectangle?

The diameter is one diagonal of the rectangle.

Pythagoras: 130² = x² + (2.4x)² = 6.76x², therefore x² = 2500 m, therefore x (width) = 50 m.

Rectangle area = 50 x 2.4 x 50 = 6,000 m²

Disk area = pi x (130/2)² = 13,273 m² approximately.

Therefore the difference area = 7,273 m² approximately.

Edited by ant0n
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33 minutes ago, ant0n said:

The diameter is one diagonal of the rectangle.

Pythagoras: 130² = x² + (2.4x)² = 6.76x², therefore x² = 2500 m, therefore x (width) = 50 m.

Rectangle area = 50 x 2.4 x 50 = 6,000 m²

Disk area = pi x (130/2)² = 13,273 m² approximately.

Therefore the difference area = 7,273 m² approximately.

Nice to see you back, Dan.

Correct again, Anton.

It's back to you for another problem.

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Great

While in jail, John was realizing that if he had 3 cigarette ends, he could mix them together to get a full cigarette. He got a cigarette butt each time after smoking a full cigarette.

One day, in his cell, John had a conversation with one of his two cellmates, Jim : « I’d really like to smoke 5 cigarettes today but I’ve got only 10 butts. I guess they won’t be enough. »

-        Jim asked: "Why don’t you steal some butts from our cellmate Mike?", pointing at that little pile of butts on their cellmate’s bed. Mike was out all day because of community service.

-         - I can’t, John replied. « Mike always carefully counts his remaining butts and he would literally kill me if he noticed I stole some. »

However, upon reflexion, John realized there actually was a way to smoke 5 cigarettes without irritating Mike. What was his plan?

Edited by ant0n
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15 hours ago, ant0n said:

Great

While in jail, John was realizing that if he had 3 cigarette ends, he could mix them together to get a full cigarette. He got a cigarette butt each time after smoking a full cigarette.

One day, in his cell, John had a conversation with one of his two cellmates, Jim : « I’d really like to smoke 5 cigarettes today but I’ve got only 10 butts. I guess they won’t be enough. »

-        Jim asked: "Why don’t you steal some butts from our cellmate Mike?", pointing at that little pile of butts on their cellmate’s bed. Mike was out all day because of community service.

-         - I can’t, John replied. « Mike always carefully counts his remaining butts and he would literally kill me if he noticed I stole some. »

However, upon reflexion, John realized there actually was a way to smoke 5 cigarettes without irritating Mike. What was his plan?

My solution:

John has 10 butts, that will allow him to make 3 cigarettes with 1 butt over.

He smokes those 3 cigarettes giving him 3 new butts plus the one he already had = 4 butts.

He makes a fourth cigarettes using 3 of these 4 butts and smokes it , leaving him with 2 butts.

To make a fifth cigarette he borrows a butt from his cellmate's pile. That allows him to roll and smoke a fifth cigarette. When he is finished he has a butt left which he returns to his cellmates pile leaving him non the wiser.

Job done!

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6 minutes ago, Ozymandias said:

My solution:

John has 10 butts, that will allow him to make 3 cigarettes with 1 butt over.

He smokes those 3 cigarettes giving him 3 new butts plus the one he already had = 4 butts.

He makes a fourth cigarettes using 3 of these 4 butts and smokes it , leaving him with 2 butts.

To make a fifth cigarette he borrows a butt from his cellmate's pile. That allows him to roll and smoke a fifth cigarette. When he is finished he has a butt left which he returns to his cellmates pile leaving him non the wiser.

Job done!

Well done!

I was still a smoker 3 3/4 years ago and I know that butt trick. Objectively, over-recycled butts must end up stinking quite bad!

Edited by ant0n
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1 hour ago, ant0n said:

I was still a smoker 3 3/4 years ago and I know that butt trick. Objectively, recycled butts must end stinking quite bad!

I was a heavy smoker - 20 to 40 per day depending on what I was doing - until the millennium New Year's Eve, 31 Dec 1999. Haven't smoked since.

Problem:   5 numbers are generated in sequence following the same rule.

They are:      X    192    Y     108     Z

Find X, Y and Z.

Edited by Ozymandias
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3 minutes ago, Ozymandias said:

I was a heavy smoker - 20 to 40 per day depending on what I was doing - until the millennium New Year's Eve, 31 Dec 1999. Haven't smoked since.

Problem:   5 numbers are generated in sequence following the same rule.

They are:      X    192    Y     108     Z

Find X, Y and Z.

234 192 150 108 66

X            Y           Z

I guess so.

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10 minutes ago, ant0n said:

234 192 150 108 66

X            Y           Z

I guess so.

No, Anton. Not the answer I am expecting. How did you get your numbers?

PS - I pressed the 'like' button by accident on post #338. Meant to click on 'confused'.

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1 minute ago, Ozymandias said:

No, Anton. Not the answer I am expecting. How did you get your numbers?

It is not the answer you're expecting HOWEVER my sequence is valid anyway: substracting 42 each time, from left to right.

If you expect some other specific sequence, you have to add conditions.

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2 minutes ago, ant0n said:

It is not the answer you're expecting HOWEVER my sequence is valid anyway: substracting 42 each time, from left to right.

If you expect some other specific sequence, you have to add conditions.

Yes, I should have stipulated that my progression was geometric. You solved the problem on the basis that it was an arithmetic progression. So, what is X, Y and Z if the sequence is geometric? Sorry about that.

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4 minutes ago, Ozymandias said:

Yes, I should have stipulated that my progression was geometric. You solved the problem on the basis that it was an arithmetic progression. So, what is X, Y and Z if the sequence is geometric? Sorry about that.

No worries

Multiply by 3/4 each step from left to right.

256 192 144 108

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1 hour ago, ant0n said:

No worries

Multiply by 3/4 each step from left to right.

256 192 144 108

Yes, Anton. That's it: X = 256, Y = 144, and Z = 81. Your turn.

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7 minutes ago, Ozymandias said:

Yes, Anton. That's it: X = 256, Y = 144, and Z = 81. Your turn.

Great

Find the number 28 as the result of a calculation involving the numbers 2, 3, 4, and 5, each one of them being used only once (parenthese allowed).

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28   =   7 x 4

=   (10 - 3) x 4

=   [(2 x 5) - 3] x 4

=   4[2(5) - 3]

Edited by Ozymandias
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11 minutes ago, Ozymandias said:

28   =   7 x 4

=   (10 - 3) x 4

=   [(2 x 5) - 3] x 4

=   4[2(5) - 3]

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Another geometry/trigonometry problem ...

Problem: A circle of radius 10cm is inscribed inside a right-angled triangle whose base length is 35cm (see diagram). What is the area coloured black?

Edited by Ozymandias
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I hope I didn't make mistakes...

- angle DGH = 90° therefore DGE = EGH = (360 - 90)/2 = 135°

- EG = GD = 10 cm

- angle FGE = 135/2 = 67.5° = FGD

- angle FEG = 180 - ( 90 - 67.5) = 22.5° = FDG

- cos22.5° = EF/10 = 0.9239 therefore EF = 9.24 cm = FD

- triangle AFD: sin(FAD) = FD/35 = 9.24/35 = 0.264 therefore angle FAD = 15.31° = FAE

- cos(FAD) = cos15.31° = AF/AD therefore AF = 0.965 x (35 - 10) = 24.13 cm

- area AFD = 1/2 x AF x FD therefore area AED = 24.13 x 9.24 = 222.96 cm²

- area FEG = 1/2 x FG x EF; sin(FEG) = FG/10 = sin22.5° therefore FG = 3.83 cm

therefore area FEG + area FDG = 2 x 1/2 x 10 x 3.83 = 38.3 cm²

- sector of the disk with angle EGD = pi x 10² x 135 / 360 = 117.81 cm²

- portion of disk EFD = 117.81 - 38.3 = 79.51 cm²

- black part = 222.96 - 79.51 = 143.42 cm²

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3 hours ago, ant0n said:

I hope I didn't make mistakes...

- angle DGH = 90° therefore DGE = EGH = (360 - 90)/2 = 135°

- EG = GD = 10 cm

- angle FGE = 135/2 = 67.5° = FGD

- angle FEG = 180 - ( 90 - 67.5) = 22.5° = FDG

- cos22.5° = EF/10 = 0.9239 therefore EF = 9.24 cm = FD

- triangle AFD: sin(FAD) = FD/35 = 9.24/35 = 0.264 therefore angle FAD = 15.31° = FAE

- cos(FAD) = cos15.31° = AF/AD therefore AF = 0.965 x (35 - 10) = 24.13 cm

- area AFD = 1/2 x AF x FD therefore area AED = 24.13 x 9.24 = 222.96 cm²

- area FEG = 1/2 x FG x EF; sin(FEG) = FG/10 = sin22.5° therefore FG = 3.83 cm

therefore area FEG + area FDG = 2 x 1/2 x 10 x 3.83 = 38.3 cm²

- sector of the disk with angle EGD = pi x 10² x 135 / 360 = 117.81 cm²

- portion of disk EFD = 117.81 - 38.3 = 79.51 cm²

- black part = 222.96 - 79.51 = 143.42 cm²

Again, Anton. Not the result I was expecting. I think, at the very beginning of your solution, your statement that the angle DGE = EGH = (360 - 90)/2 = 135° is not correct.

Using your diagram the 'black' area is equal to twice the area of the triangle ADG (ADG + AEG) minus the circle sector swept out by the angle DGE.

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8 hours ago, Ozymandias said:

Again, Anton. Not the result I was expecting. I think, at the very beginning of your solution, your statement that the angle DGE = EGH = (360 - 90)/2 = 135° is not correct.

Using your diagram the 'black' area is equal to twice the area of the triangle ADG (ADG + AEG) minus the circle sector swept out by the angle DGE.

Oh... I thought DGE = EGH... I assumed it was the result of a general law for an inscribed circle in a rectangle triangle...

Here, B1IC1 = 90° but B1IA1 =/= A1IC1 indeed:

I'm trying to find another solution.

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3 hours ago, ant0n said:

B1IC1 = 90° but B1IA1 =/= A1IC1 indeed:

This is correct. They are nor equal. They would only be equal if the sides AB and BC were equal.

3 hours ago, ant0n said:

The area required is the are of A1-C-B1-I-A1 (an area twice the size of the triangle A1IB1) less the area of the circle sector swept out by the angle B1IA1.

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The area required is the are of A-D-C-B-A (an area twice the size of the triangle ABC = ADC) less the area of the circle sector swept out by the angle DCA.

Angle CAD = arctan(10/25) = 21.8deg  >>  angle ACD = angle ACB = 90deg - 21.8deg = 68.2deg

Angle DCB =  angle ACD + angle ACB = 2(68.2deg) = 136.4deg

Area of Circle Sector swept out by the angle DCB = (136.4/360)(pi)(10^2) = 119 cm^2

Area of triangle ADC = (1/2)(base)(perp.height) = 1/2(25)(10) = 125cm^2

Area of A-D-C-B-A = 2(area triangle ADC) = 2(125cm^2) = 250cm^2

'Black' area = 250cm^2 - 119cm^2 = 131cm^2

Edited by Ozymandias
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17 minutes ago, Ozymandias said:

The area required is the are of A-D-C-B-A (an area twice the size of the triangle ABC = ADC) less the area of the circle sector swept out by the angle DCA.

Angle CAD = arctan(10/25) = 21.8deg  >>  angle ACD = angle ACB = 90deg - 21.8deg = 68.2deg

Angle DCB =  angle ACD + angle ACB = 2(68.2deg) = 136.4deg

Area of Circle Sector swept out by the angle DCB = (136.4/360)(pi)(10^2) = 119 cm^2

Area of triangle ADC = (1/2)(base)(perp.height) = 1/2(25)(10) = 125cm^2

Area of A-D-C-B-A = 2(area triangle ADC) = 2(125cm^2) = 250cm^2

'Black' area = 250cm^2 - 119cm^2 = 131cm^2

Oh no, I was just having the solution. You revealed it too early.

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18 minutes ago, Ozymandias said:

The area required is the are of A-D-C-B-A (an area twice the size of the triangle ABC = ADC) less the area of the circle sector swept out by the angle DCA.

Angle CAD = arctan(10/25) = 21.8deg  >>  angle ACD = angle ACB = 90deg - 21.8deg = 68.2deg

Angle DCB =  angle ACD + angle ACB = 2(68.2deg) = 136.4deg

Area of Circle Sector swept out by the angle DCB = (136.4/360)(pi)(10^2) = 119 cm^2

Area of triangle ADC = (1/2)(base)(perp.height) = 1/2(25)(10) = 125cm^2

Area of A-D-C-B-A = 2(area triangle ADC) = 2(125cm^2) = 250cm^2

'Black' area = 250cm^2 - 119cm^2 = 131cm^2

Why give a hint if you reveal it all 11 minutes later? I was about to typewrite my solution when I realized you revealed it all 5 minutes earlier.