Easy math puzzles
Poll
 | 21 votes (45.65%) | ||
| | 14 votes (30.43%) | ||
| | 6 votes (13.04%) | ||
| | 3 votes (6.52%) | ||
| | 12 votes (26.08%) | ||
| | 3 votes (6.52%) | ||
| | 6 votes (13.04%) | ||
| | 5 votes (10.86%) | ||
| | 12 votes (26.08%) | ||
| | 10 votes (21.73%) |
46 members have voted
November 28th, 2024 at 2:25:03 PM
permalink
Happy Thanksgiving my fellow math nerds! Here is an easy geometry puzzle to help celebrate.
ACED is a square of side length 2. AB = BC. What is the area of the blue region?
I label some extra points for purposes of discussion.
ACED is a square of side length 2. AB = BC. What is the area of the blue region?
I label some extra points for purposes of discussion.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 28th, 2024 at 5:22:52 PM
permalink
I've constructed a few more lines.
By similar triangles JKB has sides in the ratio of 2 to 1 (ref BCE), and JKH has equal sides (ref CBH), thus one can say that JK=KH=x and KB=2x. But BH=3x=1. So x=1/3.
So area of JKH = 1/3 * 1/3 * 1/2 = 1/18.
Area of LKB is twice this = 2/18.
So total of the four triangles that make up BGHJ = 6/18 = 1/3.
Another way of looking it is consider copying move BGK to the right of BKJ and similarly KGH moves, to form the rectangle. This rectangle has height of 1 (BH) and width = 1/3; hence the area = 1/3.
By similar triangles JKB has sides in the ratio of 2 to 1 (ref BCE), and JKH has equal sides (ref CBH), thus one can say that JK=KH=x and KB=2x. But BH=3x=1. So x=1/3.
So area of JKH = 1/3 * 1/3 * 1/2 = 1/18.
Area of LKB is twice this = 2/18.
So total of the four triangles that make up BGHJ = 6/18 = 1/3.
Another way of looking it is consider copying move BGK to the right of BKJ and similarly KGH moves, to form the rectangle. This rectangle has height of 1 (BH) and width = 1/3; hence the area = 1/3.
November 28th, 2024 at 8:19:52 PM
permalink
Quote: charliepatrickI've constructed a few more lines.
By similar triangles JKB has sides in the ratio of 2 to 1 (ref BCE), and JKH has equal sides (ref CBH), thus one can say that JK=KH=x and KB=2x. But BH=3x=1. So x=1/3.
So area of JKH = 1/3 * 1/3 * 1/2 = 1/18.
Area of LKB is twice this = 2/18.
So total of the four triangles that make up BGHJ = 6/18 = 1/3.
Another way of looking it is consider copying move BGK to the right of BKJ and similarly KGH moves, to form the rectangle. This rectangle has height of 1 (BH) and width = 1/3; hence the area = 1/3.
link to original post
I agree! Here is my solution (PDF).
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 29th, 2024 at 4:31:05 AM
permalink
Two circles are inscribed in a rectangle of height 81. There is a line segment of length 56 extending to the edge of both circles and goes through where the circles meet and is parallel to the vertical edge of the rectangle.
How wide is the rectangle?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 29th, 2024 at 6:34:23 AM
permalink
Conditions: AB=BC=81, EF=56, EF//AD//BC, Line 'EF' passes through the contact point where the two circles meet.
Solution: Join the centre of both circles. Let the centre of the small circle be C1 and the centre of the large circle be C2. Let the radius of the small circle be R1 and the radius of the large circle be R2. Call the intersection point M.
Draw a line from C1 to meet EF at right angles and call this point X. Draw a line from C2 to meet EF at right angles and call this point Y.
By the law of triangles XY = EF/2 = 56/2 = 28.
By deduction: radius of large circle (R2) = (81 - 28 - R1) = 53 - R1
Distance C1-C2 = 53 - R1 + R1 = 53.
Extend line C2-Y to AD and draw a line from C1 parallel with AD to meet DC. Call the intersection point of the two construction lines point Z.
By deduction: C1-Z = 28,
By deduction: C2-Z = Sq, root (53*53 - 28*28) = 45
Answer: CD = R1 + 45 + 53 - R1
= 98
Solution: Join the centre of both circles. Let the centre of the small circle be C1 and the centre of the large circle be C2. Let the radius of the small circle be R1 and the radius of the large circle be R2. Call the intersection point M.
Draw a line from C1 to meet EF at right angles and call this point X. Draw a line from C2 to meet EF at right angles and call this point Y.
By the law of triangles XY = EF/2 = 56/2 = 28.
By deduction: radius of large circle (R2) = (81 - 28 - R1) = 53 - R1
Distance C1-C2 = 53 - R1 + R1 = 53.
Extend line C2-Y to AD and draw a line from C1 parallel with AD to meet DC. Call the intersection point of the two construction lines point Z.
By deduction: C1-Z = 28,
By deduction: C2-Z = Sq, root (53*53 - 28*28) = 45
Answer: CD = R1 + 45 + 53 - R1
= 98
Casino Enemy No.1
November 30th, 2024 at 6:30:36 AM
permalink
Quote: davethebuilder
Answer: CD = R1 + 45 + 53 - R1
= 98
link to original post
I agree.
Dave, can you please put answers in the future in spoiler tags.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 30th, 2024 at 8:01:18 AM
permalink
No problem, will do.
By the way, there's a typo in the first line, it should read AD=BC=81.
By the way, there's a typo in the first line, it should read AD=BC=81.
Casino Enemy No.1
December 11th, 2024 at 7:26:44 PM
permalink
Here is an easy puzzle that came up on my Facebook feed.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 11th, 2024 at 8:59:36 PM
permalink
Casino Enemy No.1
December 12th, 2024 at 4:06:39 AM
permalink
Quote: davethebuilderLet the height of the table be ‘x’ Let the height of the Cat be ‘y’ Let the height of the Tortoise be ‘z’ By deduction: x = 170 – y + z By deduction: x = 130 + y – z Solving: y = z + 20 Substituting: x = 170 – (z + 20) + z Answer: x = 150cm Check: x = 130 + (z + 20) – z Answer: x = 150cm
link to original post
I agree.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 12th, 2024 at 4:34:36 AM
permalink
The annual GCHQ puzzle appeared yesterday, I don't know the answers but they were in the paper today.
You can download the PDF here: https://www.gchq.gov.uk/news/gchq-christmas-challenge-2024 or https://www.gchq.gov.uk/files/GCHQ%20Christmas%20Challenge%202024.pdf
Edit added SPOILER : I've found, but not read, the answers can be found here : https://www.pressreader.com/uk/daily-mail/20241212/282093462326112
You can download the PDF here: https://www.gchq.gov.uk/news/gchq-christmas-challenge-2024 or https://www.gchq.gov.uk/files/GCHQ%20Christmas%20Challenge%202024.pdf
Edit added SPOILER : I've found, but not read, the answers can be found here : https://www.pressreader.com/uk/daily-mail/20241212/282093462326112
December 12th, 2024 at 2:10:04 PM
permalink
Sorry to step on the puzzle above. For those more inclined to geometry, here is one for you.
The yellow circle has a radius of 1. What is the area of the red circle?
The yellow circle has a radius of 1. What is the area of the red circle?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 12th, 2024 at 3:53:20 PM
permalink
Let C be the height of the cat, T the height of the turtle, and X the height of the table, which is the desired value
Top image: C + X - T = 170
Bottom image: T + X - C = 130
Add them up: (C - C) + 2X + (T - T) = 300
2X = 300, so X = the height of the table = 150 cm
December 12th, 2024 at 4:14:19 PM
permalink
Quote: charliepatrickThe annual GCHQ puzzle appeared yesterday, I don't know the answers but they were in the paper today.
link to original post
I got into a bit of a discussion about this over on the Straight Dope Message Board:
Santa's trail says, "We Wish You a Merry Christmas" twice, in Morse code - but since it's British (especially as puzzle 2 expects you to know what a Blue Peter Badge and Blackpool Tower are), shouldn't it be, "Happy Christmas"?
December 12th, 2024 at 6:02:45 PM
permalink
Quote: WizardThe yellow circle has a radius of 1. What is the area of the red circle?
link to original post
Let Y be the center of the yellow circle, R the center of the red circle, B the center of the blue semicircle, and P and Q the points of tangency of the yellow circle to the top and right edges of the bounding square.
Let x be the radius of the red circle, and y the radius of the yellow semicircle.
Since the distance from the center of the circle to Q is 1, the distance from P to the upper right corner of the square is also 1, so the distance from P to B is 1 - y.
Also, since the yellow and blue circles are tangent, YB = the sum of their radii = 1 + y.
Pythagorean Theorem: 1^2 + (1 - y)^2 = (1 + y)^2
y = 1/4
Draw a line through R that is perpendicular to the top side (and parallel to the right side) of the square.
Let C be the point where this line intersects the top of the square.
Since the line segment from R to the right edge of the square is a radius of R, its distance is x, which means the distance from C to the upper right corner of the square is also x, and CB has length y - x. Also, since the blue semicircle and red circle are tangent, RB = y + x.
Pythagorean Theorem: (y - x)^2 + z^2 = (y + x)^2, or z^2 = (y + x)^2 - (y - x)^2 = 4xy = x (since y = 1/4).
CR = sqrt(x)
Let D be the point where YQ intersects the line through R parallel to the right edge; RD = 1 - sqrt(x), and YD = 1 - x
Since the yellow and red circles are tangent, YR = 1 + x
Pythagorean Theorem: (1 - sqrt(x))^2 + (1 - x)^2 = (1 + x)^2 = 4x
Take the square root of both sides: 1 - sqrt(x) = 2 sqrt(x), or 1 = 3 sqrt(x)
sqrt(x) = 1/3, so the radius of the red circle = (1/3)^2 = 1/9.
December 12th, 2024 at 7:24:33 PM
permalink
December 13th, 2024 at 4:08:35 AM
permalink
Don & CD -- I agree!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 13th, 2024 at 6:59:05 AM
permalink
Quote: WizardDon & CD -- I agree!
link to original post
But ChesterDog gets the beer this time because you did ask for the area, and my answer was the radius.
December 13th, 2024 at 7:17:56 AM
permalink
^ When it comes to beer, it's the volume that really counts. Luckily we still have pints here!
December 15th, 2024 at 5:14:25 PM
permalink
Here is another one that came through my Facebook feed:
x^2 - y^2 = 64
xy = 8
Find x+y.
x^2 - y^2 = 64
xy = 8
Find x+y.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 15th, 2024 at 7:05:29 PM
permalink
Quote: WizardHere is another one that came through my Facebook feed:
x^2 - y^2 = 64
xy = 8
Find x+y.
link to original post
x^2 - (8/x)^2 = 64
x^2 - 64 / x^2 = 64
(x^2)^2 - 64 = 64 x^2
(x^2)^2 - 64 x^2 - 64 = 0
(x^2)^2 - 64 x^2 + 32^2 = 32 * 34 = 64 * 17
(x^2 - 32)^2 = (8 sqrt(17))^2
x^2 = 32 +/- 8 sqrt(17)
y^2 = x^2 - 64 = +/- 8 sqrt(17) - 32
x^2 + y^2 = 16 sqrt(17)
(x + y)^2 = x^2 + 2xy + y^2 = x^2 + y^2 + 16 = 16 + 16 sqrt(17)
x + y = +/- 4 sqrt(1 + sqrt(17))
I know the positive is correct; not sure about the negative
December 15th, 2024 at 8:51:40 PM
permalink
Quote: ThatDonGuy
x^2 - (8/x)^2 = 64
x^2 - 64 / x^2 = 64
(x^2)^2 - 64 = 64 x^2
(x^2)^2 - 64 x^2 - 64 = 0
(x^2)^2 - 64 x^2 + 32^2 = 32 * 34 = 64 * 17
(x^2 - 32)^2 = (8 sqrt(17))^2
x^2 = 32 +/- 8 sqrt(17)
y^2 = x^2 - 64 = +/- 8 sqrt(17) - 32
x^2 + y^2 = 16 sqrt(17)
(x + y)^2 = x^2 + 2xy + y^2 = x^2 + y^2 + 16 = 16 + 16 sqrt(17)
x + y = +/- 4 sqrt(1 + sqrt(17))
I know the positive is correct; not sure about the negative
link to original post
That's not what I got, but maybe I'm in the one in error. Hopefully we get a third opinion.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 16th, 2024 at 3:35:51 AM
permalink
^ If x and y are solutions then so would be -x and -y. The square of -x = the square of x, and xy = (-x)(-y). I think I saw this puzzle on facebook.
December 16th, 2024 at 7:30:54 AM
permalink
x² - y² = 64
xy = 8
Find x + y
Solution:
If xy = 8, y = 8/x
x² - (8/x)² = 64
x² - (64/x²) = 64
(x²)² - 64x² - 64 = 0
Convert quartic equation into quadratic form by letting a = x²
a² - 64a - 64 = 0
a = [Sq. root 64 + (Sq. root (64*68))]/2
x = Sq. root {[Sq. root 64 + (Sq. root (64*68))]/2}
x = 8.061317
Substituting: y = 8/x
y = 8/8.061317
= 0.992393
Solution: x + y = 8.061317 + 0.992393
= 9.05371
Casino Enemy No.1
December 16th, 2024 at 8:24:19 AM
permalink
I concede that Dave and Don agree and calculations in Excel confirms it. I think nobody will mind if I move this out of Spoiler tags.
Using trial and error these values of x and y come close:
x = 8.061315
y = 0.992370661
xy = 7.999812497
x+y = 9.053685661
I was trying to argue why my answer was also right, but in putting it in writing for this post, I caught my error.
Using trial and error these values of x and y come close:
x = 8.061315
y = 0.992370661
xy = 7.999812497
x+y = 9.053685661
I was trying to argue why my answer was also right, but in putting it in writing for this post, I caught my error.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 24th, 2024 at 8:44:10 AM
permalink
You roll an icosahedron (20-sided die) until it lands on the same side it already landed on at any point.
1. What is the expected number of rolls, including the ending roll?
2. Is there a formula for the expected rolls for an n-sided die?
1. What is the expected number of rolls, including the ending roll?
2. Is there a formula for the expected rolls for an n-sided die?
Last edited by: Wizard on Dec 24, 2024
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 24th, 2024 at 9:13:19 AM
permalink
Quote: WizardYou roll an icosahedron (20-sided die) until it lands on the same side it already landed on.
1. What is the expected number of rolls?
2. Is there a formula for the expected rolls for an n-sided die?
link to original post
Clarification needed: does it have to be the same side as the previous roll, or do you roll until any number comes up twice (e.g. if you roll 1, 2, 4, 12, 2, you stop at the second 2)?
December 24th, 2024 at 11:37:18 AM
permalink
Quote: ThatDonGuyClarification needed: does it have to be the same side as the previous roll, or do you roll until any number comes up twice (e.g. if you roll 1, 2, 4, 12, 2, you stop at the second 2)?
link to original post
You roll until any number comes up twice. They do not need to occur on consecutive rolls. Be sure to count the final (duplicate) roll. In your example, there were 5 rolls.
p.s. I amended the original wording to be more clear.
Last edited by: Wizard on Dec 24, 2024
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 25th, 2024 at 11:51:41 AM
permalink
Quote: WizardYou roll an icosahedron (20-sided die) until it lands on the same side it already landed on at any point.
1. What is the expected number of rolls, including the ending roll?
2. Is there a formula for the expected rolls for an n-sided die?
link to original post
For N-sided dice from 2 to 20, I get these fractions:
2: 5 / 2
3: 26 / 9
4: 103 / 32
5: 2,194 / 625
6: 1,223 / 324
7: 472,730 / 117,649
8: 556,403 / 131,072
9: 21,323,986 / 4,782,969
10: 7,281,587 / 1,562,500
11: 125,858,034,202 / 25,937,424,601
12: 180,451,625 / 35,831,808
13: 121,437,725,363,954 / 23,298,085,122,481
14: 595,953,719,897 / 110,730,297,608
15: 26,649,932,810,926 / 4,805,419,921,875
16: 3,211,211,914,492,699 / 562,949,953,421,312
17: 285,050,975,993,898,158,530 / 48,661,191,875,666,868,481
18: 549,689,343,118,061 / 91,507,169,819,844
19: 640,611,888,918,574,971,191,834 / 104,127,350,297,911,241,532,841
20: 4,027,894,135,040,576,041 / 640,000,000,000,000,000
I am trying to find a formula, but so far it has eluded me
