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46 members have voted
November 28th, 2024 at 2:25:03 PM
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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
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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
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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.
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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
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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
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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
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Quote: davethebuilder
Answer: CD = R1 + 45 + 53 - R1
= 98
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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
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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