Comments for think again! http://easyquestion.net/thinkagain seductive math problems for the modern mind Thu, 11 Mar 2010 21:53:38 +0000 http://wordpress.org/?v=2.6.5 Comment on The Ice Ocean Cathedral by Richard Sabey http://easyquestion.net/thinkagain/2010/03/12/the-ice-ocean-cathedral/#comment-48143 Richard Sabey Thu, 11 Mar 2010 21:12:13 +0000 http://easyquestion.net/thinkagain/?p=2898#comment-48143 Sorry: I meant that comment "Then it wouldn’t be equilateral." to be a response to the phrase "the base is 5/6 of the height" in the statement of the problem. Sorry: I meant that comment “Then it wouldn’t be equilateral.” to be a response to the phrase “the base is 5/6 of the height” in the statement of the problem.

]]> Comment on The Ice Ocean Cathedral by Richard Sabey http://easyquestion.net/thinkagain/2010/03/12/the-ice-ocean-cathedral/#comment-48141 Richard Sabey Thu, 11 Mar 2010 21:10:46 +0000 http://easyquestion.net/thinkagain/?p=2898#comment-48141 Then it wouldn't be equilateral. Let's suppose that, like the buildings in the pictures, it's isosceles, and the sloping sides are equal. Let the base be 10x. Then the height is 12x. By Pythagoras's Theorem, each sloping side is the hypotenuse of a right-angled triangle whose legs are 5x and 12x, so it is 13x. Then the perimeter is 2*13x+10x=36x, which is 3 times the height. This perimeter is 55m, so the height is 55/3 m = 18 1/3 m. Then it wouldn’t be equilateral. Let’s suppose that, like the buildings in the pictures, it’s isosceles, and the sloping sides are equal. Let the base be 10x. Then the height is 12x. By Pythagoras’s Theorem, each sloping side is the hypotenuse of a right-angled triangle whose legs are 5x and 12x, so it is 13x. Then the perimeter is 2*13x+10x=36x, which is 3 times the height. This perimeter is 55m, so the height is 55/3 m = 18 1/3 m.

]]> Comment on The Ice Ocean Cathedral by Michael Maguire http://easyquestion.net/thinkagain/2010/03/12/the-ice-ocean-cathedral/#comment-48136 Michael Maguire Thu, 11 Mar 2010 20:38:23 +0000 http://easyquestion.net/thinkagain/?p=2898#comment-48136 Let h = height of the church. The base, therefore = 5h/6, as does each side. 3•5h/6 = the perimeter = 55 meters. 5h/2 = 55 5h = 110 h = 22 meters Let h = height of the church.
The base, therefore = 5h/6, as does each side.
3•5h/6 = the perimeter = 55 meters.
5h/2 = 55
5h = 110
h = 22 meters

]]> Comment on Just peanuts by Jan Nordgreen http://easyquestion.net/thinkagain/2010/03/11/just-peanuts/#comment-47993 Jan Nordgreen Thu, 11 Mar 2010 05:37:12 +0000 http://easyquestion.net/thinkagain/?p=2895#comment-47993 In my class room I have seen that even simple arithmetic can give multiple equally correct approaches. :) More problems will come from this Norwegian source (Norwegian is my mother tongue). Google translated it like this to English: This task is about three children - John, Peter and Kari - who share a bag of 770 peanuts. Every time John gets 4 peanuts, get Per 3 peanuts, and every time John gets 6 peanuts, get Kari 7 peanuts. How many peanuts are Ola? In my class room I have seen that even simple arithmetic can give multiple equally correct approaches. :)

More problems will come from this Norwegian source (Norwegian is my mother tongue).

Google translated it like this to English:

This task is about three children - John, Peter and Kari - who share a bag of 770 peanuts.

Every time John gets 4 peanuts, get Per 3 peanuts, and every time John gets 6 peanuts, get Kari 7 peanuts.

How many peanuts are Ola?

]]> Comment on Just peanuts by Michael Maguire http://easyquestion.net/thinkagain/2010/03/11/just-peanuts/#comment-47977 Michael Maguire Thu, 11 Mar 2010 03:53:12 +0000 http://easyquestion.net/thinkagain/?p=2895#comment-47977 MA: One of the things I love most about math beyond simple arithmetic is that there are often multiple equally correct approaches. Maybe it's because statistics has never been a strength of mine but that approach leaves me scratching my head. Out of curiosity, with your eye for statistics, do the odds of 12:9:14 just leap out at you without much effort? Maybe it's time for me to find a statistics course at the local community college. To whoever it may concern: Mattenøtter apparently has some gems in it but unfortunately I need a translator. Please feel free to post more of them! MA:

One of the things I love most about math beyond simple arithmetic is that there are often multiple equally correct approaches. Maybe it’s because statistics has never been a strength of mine but that approach leaves me scratching my head. Out of curiosity, with your eye for statistics, do the odds of 12:9:14 just leap out at you without much effort? Maybe it’s time for me to find a statistics course at the local community college.

To whoever it may concern:

Mattenøtter apparently has some gems in it but unfortunately I need a translator. Please feel free to post more of them!

]]> Comment on Just peanuts by Mike Anderson http://easyquestion.net/thinkagain/2010/03/11/just-peanuts/#comment-47923 Mike Anderson Wed, 10 Mar 2010 22:43:54 +0000 http://easyquestion.net/thinkagain/?p=2895#comment-47923 MM: Many of my students are afraid of fractions, so I rephrased the conditions as "peanut odds" for Alan:Boris:Carla. The odds are 12:9:14, comprising 35 parts in all. Then 770/35 = 22 peanuts per part, and multiplication gives the (same) answer. Nice problem! MM:

Many of my students are afraid of fractions, so I rephrased the conditions as “peanut odds” for Alan:Boris:Carla. The odds are 12:9:14, comprising 35 parts in all. Then 770/35 = 22 peanuts per part, and multiplication gives the (same) answer.

Nice problem!

]]> Comment on Just peanuts by Michael Maguire http://easyquestion.net/thinkagain/2010/03/11/just-peanuts/#comment-47918 Michael Maguire Wed, 10 Mar 2010 22:31:49 +0000 http://easyquestion.net/thinkagain/?p=2895#comment-47918 A+(3/4•A)+(7/6•A)=770 A•(1+9/12+14/12)=770 A•(35/12)=770 A=770•12/35=264 Alan gets 264. Boris gets 198. Carla gets 308. I'm going to share this with my son. Thanks for the problem! A+(3/4•A)+(7/6•A)=770
A•(1+9/12+14/12)=770
A•(35/12)=770
A=770•12/35=264

Alan gets 264.
Boris gets 198.
Carla gets 308.

I’m going to share this with my son. Thanks for the problem!

]]> Comment on Was it rigged? by Anceps http://easyquestion.net/thinkagain/2010/03/05/2844/#comment-47743 Anceps Tue, 09 Mar 2010 09:06:22 +0000 http://easyquestion.net/thinkagain/?p=2844#comment-47743 I found the same result but in a third way: The first one (A) is elected. The second one to be elected (B) has 15 chances on 19 to seat at least 3 seats away from A and 2 chances on 19 to seat exactly two seats away from A. In the first case, there's 14 (20 seats - 2 taken seats - 4 seats nearby) chances on 18 to elect C away from A and B. In the second case, there's 15 (20 - 2 taken seats - 3 seats nearby) chances on 18. Probability no one seats just by another one: 15/19 * 14 / 18 + 2/19 * 15/18 = 210/(19*18) + 30/(19*18) = 240 / (19*18) = 40 / 57. Probability at least one seats just by another: 17/57. I found the same result but in a third way:

The first one (A) is elected.
The second one to be elected (B) has 15 chances on 19 to seat at least 3 seats away from A and 2 chances on 19 to seat exactly two seats away from A.
In the first case, there’s 14 (20 seats - 2 taken seats - 4 seats nearby) chances on 18 to elect C away from A and B.
In the second case, there’s 15 (20 - 2 taken seats - 3 seats nearby) chances on 18.

Probability no one seats just by another one:
15/19 * 14 / 18 + 2/19 * 15/18
= 210/(19*18) + 30/(19*18) = 240 / (19*18) = 40 / 57.
Probability at least one seats just by another: 17/57.

]]> Comment on Never tired bouncing ball by Richard Sabey http://easyquestion.net/thinkagain/2010/03/08/never-tired-bouncing-ball/#comment-47734 Richard Sabey Tue, 09 Mar 2010 08:13:42 +0000 http://easyquestion.net/thinkagain/?p=2903#comment-47734 You say that, in your example, it /ends up/ at D. That is, once it reaches a corner, we regard its trajectory to have ended. Then it can't end up at A, for, if it did, its projection onto AB must have made an even number, 2x, of horizontal traversals (from left to right or right to left), and its projection onto AD must have made an even number, 2y, of vertical traversals (from bottom to top or top to bottom). Then halfway through its trajectory those projections had made x horizontal traversals and y vertical traversals, and so it was at a corner, so it got no further. You say that, in your example, it /ends up/ at D. That is, once it reaches a corner, we regard its trajectory to have ended. Then it can’t end up at A, for, if it did, its projection onto AB must have made an even number, 2x, of horizontal traversals (from left to right or right to left), and its projection onto AD must have made an even number, 2y, of vertical traversals (from bottom to top or top to bottom). Then halfway through its trajectory those projections had made x horizontal traversals and y vertical traversals, and so it was at a corner, so it got no further.

]]> Comment on Never tired bouncing ball by Jan Nordgreen http://easyquestion.net/thinkagain/2010/03/08/never-tired-bouncing-ball/#comment-47714 Jan Nordgreen Mon, 08 Mar 2010 23:33:10 +0000 http://easyquestion.net/thinkagain/?p=2903#comment-47714 Yes, always. Yes, always.

]]>