## Poll

18 votes (51.42%) | |||

13 votes (37.14%) | |||

5 votes (14.28%) | |||

2 votes (5.71%) | |||

11 votes (31.42%) | |||

3 votes (8.57%) | |||

6 votes (17.14%) | |||

5 votes (14.28%) | |||

10 votes (28.57%) | |||

8 votes (22.85%) |

**35 members have voted**

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

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.

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?

Quote:ThatDonGuyHere'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?

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.

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.

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/GoogolplexIn 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".[1] 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".[2] It thus became standardized to 10(

^{10100}).

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

Quote:charliepatrick36 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...

Quote:teliotLast 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.)

Quote:ThatDonGuyKeep going; you're getting there...

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.

Quote:charliepatrickQuote:ThatDonGuyHere'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

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