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teliot Joined: Oct 19, 2009
• Threads: 39
• Posts: 2180
March 7th, 2021 at 6:55:56 AM permalink
Quote: ThatDonGuy

Quote: teliot

Optimal strategy -- thanks to charliepatrick for the code.

Minimum pair (i,j) with i < j to draw a second ball to replace i.

1, 2
2, 3
3, 4
4, 5
5, 6
6, 7
7, 8
8, 9
9, 10
10, 11
11, 12
12, 13
13, 15
14, 16
15, 17
16, 18
17, 19
18, 21
19, 22
20, 23
21, 25
22, 26
23, 27
24, 29
25, 30
26, 32
27, 33
28, 35
29, 36
30, 38
31, 40
32, 42
33, 43
34, 45
35, 47
36, 49
37, 52
38, 54
39, 56
40, 59
41, 62
42, 65
43, 69
44, 73
45, 79

Return = 16.222548

I get 45,293,117,053,521 / 2,791,985,396,200, which is 16.222547981507

The strategy is to draw when A < (183 B - B^2) / 182, where A and B are the smallest and second-smallest numbers.
This is equivalent to B > 183/2 - sqrt(183^2 / 4 - 182 A) for B < 183/2.
It appears to match the above list of pairs.

Fabulous! It is surprising to me that using perfect strategy adds so little value.

Here is another math problem.

Last night on the phone my grandson who just turned six asked me the question what is the number that comes just before a googolplex. I am interested to see what responses we get here.
Last edited by: teliot on Mar 7, 2021
Poetry website: www.totallydisconnected.com
ThatDonGuy Joined: Jun 22, 2011
• Threads: 99
• Posts: 4740
March 7th, 2021 at 8:37:16 AM permalink
Here's another problem - this is one I was given when I was a freshman in high school (1977), although the original problem said there was only one solution, and I surprised the author when I came up with multiple solutions.

ABC is an isosceles triangle, with sides AB and AC the same length.
A line going through one of the three points and intersecting the opposite side of the triangle divides the triangle into two non-congruent isosceles triangles.
For what values of angle BAC is this possible?
charliepatrick Joined: Jun 17, 2011
• Threads: 33
• Posts: 2359
March 7th, 2021 at 10:42:14 AM permalink
Quote: ThatDonGuy

Here's another problem - this is one I was given when I was a freshman in high school (1977), although the original problem said there was only one solution, and I surprised the author when I came up with multiple solutions.

ABC is an isosceles triangle, with sides AB and AC the same length.
A line going through one of the three points and intersecting the opposite side of the triangle divides the triangle into two non-congruent isosceles triangles.
For what values of angle BAC is this possible?

36 or 108 degrees.

First consider a line from B intersecting AC at a point D such that BC=BD and BD=AD. From {BAC} being isosceles if <BAC=x then <DCB=(180-x)/2. From {DBC} being isosceles this means <CDB=<DCB so <DBC=x. Since {BDA} is isosceles <ABD=x. So <CBA=2x=<DCB=(180-x)/2. So x=36 degrees. Note that BDC and ABC are similar.

Another solution is a relatively flatter triangle using a line from A to D. Using similar logic the angles are 36 36 108 for ABC and ADC, with ABD being 36 72 72.

Interestingly having created a triangle from the original ABC, you can now recreate the effect in both the new triangles .ADB and BDC.
charliepatrick Joined: Jun 17, 2011
• Threads: 33
• Posts: 2359
March 7th, 2021 at 11:07:14 AM permalink
Quote: teliot

...Here is another math problem.

Last night on the phone my grandson who just turned six asked me the question what is the number that comes just before a googolplex. I am interested to see what responses we get here.

The story goes that the term was devised by the nephew of Dr Edward Kasner shortly after the term googol was devised. So timewise the answer is googol.

In the mathematical sense it depends on whether you're looking at just integers, in which case if you ordered them by value, N-1 comes just before N. I'm not sure how you order Real numbers since if you said N-x came just before N, then where does N-(x/2) come. I suppose given you've asked the question assuming there is an order to numbers, then you would mean the former; so it would be (googol-1).

References

https://www.wsj.com/articles/SB108575924921724042

Quote: https://en.wikipedia.org/wiki/Googolplex

In 1920, Edward Kasner's nine-year-old nephew, Milton Sirotta, coined the term googol, which is 10100, and then proposed the further term googolplex to be "one, followed by writing zeroes until you get tired". Kasner decided to adopt a more formal definition because "different people get tired at different times and it would never do to have Carnera a better mathematician than Dr. Einstein, simply because he had more endurance and could write for longer". It thus became standardized to 10(10100).

chevy Joined: Apr 15, 2011
• Threads: 2
• Posts: 84
March 7th, 2021 at 11:24:51 AM permalink

I can only get one???
Let a line through B hit AC at point D

x=angle BAC=BAC (for short)
y=ABC=BCA

So for triangle ABC
x+2y=180

BDC=BCA=y Since smaller triangle BCD is isosceles
So BDA=180-y

DAB=DBA=x Since smaller triangle BAD is isosceles
So for triangle BAD

2x+(180-y)=180
y=2x

Back to triangle ABC
x+2y=180
x+2(2x)=180

x=36. (angle BAC)

Since line through B or line through C is same....this is only one I get.

For line through A, I would get BAC = 90, but that would divide the triangle into two identical (i.e congruent) smaller triangles.

So I am missing the other option you found.

EDIT: I see charliepatrick found other answer

ThatDonGuy Joined: Jun 22, 2011
• Threads: 99
• Posts: 4740
March 7th, 2021 at 12:35:28 PM permalink
Quote: charliepatrick

36 or 108 degrees.

First consider a line from B intersecting AC at a point D such that BC=BD and BD=AD. From {BAC} being isosceles if <BAC=x then <DCB=(180-x)/2. From {DBC} being isosceles this means <CDB=<DCB so <DBC=x. Since {BDA} is isosceles <ABD=x. So <CBA=2x=<DCB=(180-x)/2. So x=36 degrees. Note that BDC and ABC are similar.

Another solution is a relatively flatter triangle using a line from A to D. Using similar logic the angles are 36 36 108 for ABC and ADC, with ABD being 36 72 72.

Interestingly having created a triangle from the original ABC, you can now recreate the effect in both the new triangles .ADB and BDC.

Keep going; you're getting there...
ThatDonGuy Joined: Jun 22, 2011
• Threads: 99
• Posts: 4740
March 7th, 2021 at 12:41:22 PM permalink
Quote: teliot

Last night on the phone my grandson who just turned six asked me the question what is the number that comes just before a googolplex. I am interested to see what responses we get here.

I've answered this one before...

It's also nine times the googol'th repunit. The nth repunit is the n-digit positive integer all of whose digits are 1s. (The second repunit is 11, the third is 111, and so on.)
charliepatrick Joined: Jun 17, 2011
• Threads: 33
• Posts: 2359
March 7th, 2021 at 1:19:19 PM permalink
Quote: ThatDonGuy

Keep going; you're getting there...

Obviously you can cut a right angle triangle in half, but then the two triangles formed are congruent, assumed this wasn't a solution.

Rather than start with the triangle to be cut into half, I started with a resultant isosceles triangle and worked backwards trying to extend one of its sides to create another isosceles triangle. I looked at the ratio of angles and started with the base angles at 2x and top angle nx; using n=3 led to another answer that was <BAC=180/7.
ThatDonGuy Joined: Jun 22, 2011
• Threads: 99
• Posts: 4740
Thanks for this post from: March 15th, 2021 at 2:02:23 PM permalink
I never did post a solution to this...
Quote: charliepatrick

Quote: ThatDonGuy

Here's another problem - this is one I was given when I was a freshman in high school (1977), although the original problem said there was only one solution, and I surprised the author when I came up with multiple solutions.

ABC is an isosceles triangle, with sides AB and AC the same length.
A line going through one of the three points and intersecting the opposite side of the triangle divides the triangle into two non-congruent isosceles triangles.
For what values of angle BAC is this possible?

36 or 108 degrees.

First consider a line from B intersecting AC at a point D such that BC=BD and BD=AD. From {BAC} being isosceles if <BAC=x then <DCB=(180-x)/2. From {DBC} being isosceles this means <CDB=<DCB so <DBC=x. Since {BDA} is isosceles <ABD=x. So <CBA=2x=<DCB=(180-x)/2. So x=36 degrees. Note that BDC and ABC are similar.

Another solution is a relatively flatter triangle using a line from A to D. Using similar logic the angles are 36 36 108 for ABC and ADC, with ABD being 36 72 72.

Interestingly having created a triangle from the original ABC, you can now recreate the effect in both the new triangles .ADB and BDC.

Obviously you can cut a right angle triangle in half, but then the two triangles formed are congruent, assumed this wasn't a solution.

Rather than start with the triangle to be cut into half, I started with a resultant isosceles triangle and worked backwards trying to extend one of its sides to create another isosceles triangle. I looked at the ratio of angles and started with the base angles at 2x and top angle nx; using n=3 led to another answer that was <BAC=180/7.

That's all of them

Let ABC be an isosceles triangle with AB = AC
Let t be the measure of ABC (as well as ACB)

Case 1: D is on BC, forming triangles ABD and ACD
(a) If ABD = ADB, then ADB = t -> ADC = 180 - t -> CAD = 180 - (t + (180 - t)) = 0
(b) If ABD = BAD, then ADB = 180 - 2t -> ADC = 2t -> CAD = 180 - 3t
(b1) ADC = ACD -> 2t = t -> t = 0
(b2) ADC = CAD -> 2t = 180 - 3t -> t = 36 -> BAC = 108
108 is a solution: {108, 36, 36} -> {36, 36, 108} {72, 72, 36}
(b3) ACD = CAD -> t = 180 - 3t -> t = 45 -> ADB = ADC = 90 and BAD = CAD = 45 -> BAD and CAD are ASA-congruent
(c) If ADB = BAD, then ADB = 90 - t/2 -> ADC = 180 - (90 - t/2) = 90 + t/2 -> CAD = 180 - t - (90 + t/2) = 90 - 3/2 t
(c1) ADC = ACD -> 90 + t/2 = t -> t = 180
(c2) ADC = CAD -> 90 + t/2 = 90 - 3/2 t -> t = 0
(c3) ACD = CAD -> t = 90 - 3/2 t -> t = 36; this is the same as (b2)

Case 2: D is on AC, forming triangles ABD and CBD
Let p be the measure of ABD; CBD = t - p
Since BAC = 180 - 2t, ADB = 180 - p - (180 - 2t) = 2t - p
Since BCA = t, BDC = 180 - t - (t - p) = 180 - 2t + p
(a) If CBD = BCD, then t = t - p -> p = 0
(b) If CBD = CDB, then t - p = 180 + p - 2t -> p = 3/2 t - 90
BDA = 2t - (3/2 t - 90) = t/2 + 90; ABD = 3/2 t - 90
(b1) If BAD = ABD, then 180 - 2t = 3/2 t - 90 -> t = 540/7 -> p = 180/7
180/7 is a solution: {180/7, 540/7, 540/7} -> {180/7, 180/7, 900/7}, {360/7, 360/7, 540/7}
(b2) If BAD = BDA, then 180 - 2t = t/2 + 90 -> 90 = 5/2 t -> t = 36 -> p < 0
(b3) If ABD = BDA, then 3/2 t - 90 = t/2 + 90 -> t = 180
(c) If BCD = CDB, then t = 180 + p - 2t -> p = 3t - 180
BDA = 2t - (3t - 180) = 180 - t; ABD = 3t - 180
(c1) If BAD = ABD, then 180 - 2t = 3t - 180 -> t = 72 -> p = 36
36 is a solution: {36, 72, 72} -> {36, 108, 36} {36, 72, 72}
(c2) If BAD = BDA, then 180 - 2t = 180 - t -> t = 0
(c3) If ABD = BDA, then 3t - 180 = 180 - t -> t = 90 -> BAC = 180 - 2t = 0

Koi Joined: Feb 24, 2021
• Threads: 0
• Posts: 3
March 17th, 2021 at 6:07:20 PM permalink
First time poster to the forum.
Here is a nice algebra problem to chew on.
Solve the following system of equations for a, b, c, d, e, and f in real numbers other than the trivial solution of a = b = c = d = e = f = 0.
a + b + c + d + e = f^2
a + b + c + d + f = e^2
a + b + c + e + f = d^2
a + b + d + e + f = c^2
a + c + d + e + f = b^2
b + c + d + e + f = a^2

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