Easy math puzzles

Quote: teliot

Okay, which leads me to a cute little theorem-ette that is on the slightly non-trivial side (but not that non-trivial), related to the 153 question.

Let N > 0 be any integer. Let S be the sum of the cubes of its digits. Show that the difference (N-S) is always divisible by 3.

For example, N = 1263. S = 1^3 + 2^3 + 6^3 + 3^3 = 1 + 8 + 216 + 27 = 252. N - S = 1011= 3*337.

Quote: charliepatrick

Show Spoiler

We can split any number into its constituent digits and consider each one in turn. So a number such as 543 look at the difference between 500 and 5³, 40 and 4³, 3 and 3³. If each of these is divisible by 3, then the total will.

So the approach is to show it for one digit numbers, then two digit numbers, etc.

Consider 0<N<10. N³-N = N * (N²-1) = N * (N-1) * (N+1). One these must be a multiple of 3.

Now consider a number <100 which is 10M+N. Then we would like M³-10M to be a multiple of 3. This is the same as (M³-M)-9*M. The former (M³-M) is divisible by 3 (as per above) and 9*M is also divisible by 3. Hence the difference is divisible by 3. This proves the case for two digit numbers.

Similar logic will apply to any other digit in the number since it will split into (L³-L)-999....999*L.

Hence it applies to all integers.

Quote: unJon

Did it quick in my head but looks like:

Show Spoiler

153

Quote: teliot

Show Spoiler

153 -- which, coincidentally, is the smallest integer > 1 that is the sum of the cubes of its digits. 153 = 1^3 + 5^3 + 3^3. There are three more positive integers > 1 with this property. Bonus easy Monday question -- What are they?

Bonus bonus question -- how many fish did Simon Peter catch (New Testament)?

Quote: ThatDonGuy

Detailed Answer

By 891, one of 1, 8, 9 is in the code, but by 849, neither 8 nor 9 are in the code, so 1 is in the code, but not the third digit
By 317, 1 is not the second digit; therefore, 1 is the first digit
By 793, either 9 is the second digit or 3 is the third digit, but by 849, 9 is not in the code, so 3 is the third digit
By 725, either 7 or 5 is the second digit, but by 793, since 3 is in the code, 7 is not, so 5 is the second digit
The code is 153

joke

Quote: unJon

Did it quick in my head but looks like:

Show Spoiler

153

Quote: teliot

...Hmmm .... not sure I like the "similar logic will apply" part. Can you formalize that?

Quote: Wizard

Four congruent semicircles are arranged in a square. The radius of each circle is 1, what is the length of a side of the square?

Quote: ChesterDog

Show Spoiler

Call the centers of the topmost and rightmost circles O and P, respectively. And let point A be the upper right corner of the square.

Line segment OP is the hypotenuse of triangle OAP. OP = 2 and AP = 1. Therefore, OA = (2²-1²)^1/2 = 3^1/2.

Add 1 to the above length to get the side of the square: 3^1/2 + 1

Quote: charliepatrick

Show Spoiler

Earlier the proof was given for (X-X³) would be a multiple of 3. Now to show it applies to all larger integers.

Every integer (ABC....Z) can be considered A*10ⁿ+B*10^(n-1)...Z. The sum of the integer cubes is A³+B³+...+Z³.

Consider the A part. The difference = A*10ⁿ-A³ = A*(10ⁿ-1)+A-A³. The first part, being 999....999 is divisible by 3. The second part, by previous proof, is divisible by 3.

Hence A*10ⁿ-A³ is divisible by 3.

Since this would apply to all the other digits in any number (B,C, .... Z); the condition is true for all integers.

Eliot' Proof

Quote: teliot

Okay, which leads me to a cute little theorem-ette that is on the slightly non-trivial side (but not that non-trivial), related to the 153 question.

Let N > 0 be any integer. Let S be the sum of the cubes of its digits. Show that the difference (N-S) is always divisible by 3.

For example, N = 1263. S = 1^3 + 2^3 + 6^3 + 3^3 = 1 + 8 + 216 + 27 = 252. N - S = 1011= 3*337.

Quote: chevy

Comment for ThatDonGuy

If I am following your argument, it seems you are saying ABC has a side of 4?

You have FB=1 (I agree) and the diameter of the circle is 2. So doesn't ABC has HEIGHT of 3, then side of 2sqrt(3) from 30-60-90 triangle???

If so, then your length ratio is (2sqrt(3)/3) / (2sqrt(3)) = 1/3.
Area scales as 1/9.

If so, then it agrees with me and Wiz, if not, I am missing something in your construction.

Quote: chevy

my first pass

1) Let red circle have radius R, thus area (pi*R^2)
2) Name some points
A = top of triangle, C = center of red circle, D = center of bottom side of triangle. ACD is a line and bisects the 60 degree angle of the triangle.
B = intersection of ACD with red circle (nearest A)
D= center of left side of triangle.

3) ACD is right triangle and
CD/(AB+BC) = R/(AB+R) = sin30=.5
So AB=R

4)Height of triangle = AB+BC+CD=R+R+R=3R

5) So for a triangle of height 3R, we can fit a circle of size R

6) Draw horizontal tangent to red circle at B (top tip of the triangle). That forms a triangle with height AB=R, so the first inscribed circle must have radius R/3. Then the next triangular tip has height R/3, and inscribed circle has height R/9........

So the set of smaller circles on the top have area
pi*(R/3)^2 + pi*(R/9)^2+....geometric series = pi*(R/3)^2 / (1-1/9) = (pi*R^2)/8

So the three sets of circles have area (3/8) (pi*R^2) = 3/8 Area of red circle

Quote: Wizard

Wiz answer #2

3/8.

I see I agree with Chevy. If we're both right, please give him credit for being first.

Good problem.

Official Solve

Quote: teliot

Ok, here is a really easy follow-up question. The list in the spoiler gives all six solutions to "sum of the cubes of the digits equals the number."

My follow-up question is to find all integers that minimize the difference when it is greater than 0, that is:

minimize |(sum of cube of digits of number) - number| > 0

There are six solutions when this difference is equal to 0. I found 12 solutions to the next smallest difference.

Quote: teliot

I have a follow-up question to my follow-up question. And I don't know the answer to this.

We proved above that if N > 0 is an integer and S is the sum of the cubes of its digits, then the difference |N - S| is always divisible by 3. In my computer investigations, I've noticed that every multiple of 3 appears as the difference, at least as far up as I checked. I don't know if the following is true or false:

True or False? Let D >= 0 be divisible by 3. Then there is a positive integer N such that |N -S| = D.

Any takers?

Quote: charliepatrick

Show Spoiler

False - try finding 651, 681, 909 etc.

The simplest method is to consider that any six-digit numbers can have a maximum sum of cubes of 6*9³=4374, so the difference must exceed 95626. For longer numbers the difference just keeps getting bigger.

Thus if you haven't found all the four digit numbers by then, you never will.

Brute force shows there are several values that haven't been found (assuming my coding is correct!)

Quote: ThatDonGuy

Here's one:

There's a new game that consists of a keno blower with 100 balls, each numbered 1 through 100. Ten balls are drawn at random, and you win $1 multiplied by the lowest of the ten numbers. For example, if the balls 2, 8, 14, 15, 27, 45, 54, 62, 89, and 99 were drawn, you would win $2, as 2 is the lowest number.

What is the expected return (i.e. the mean value) of the game?

Quote: ThatDonGuy

There's a new game that consists of a keno blower with 100 balls, each numbered 1 through 100. Ten balls are drawn at random, and you win $1 multiplied by the lowest of the ten numbers. For example, if the balls 2, 8, 14, 15, 27, 45, 54, 62, 89, and 99 were drawn, you would win $2, as 2 is the lowest number.

What is the expected return (i.e. the mean value) of the game?

Quote: ThatDonGuy

Here's one:

There's a new game that consists of a keno blower with 100 balls, each numbered 1 through 100. Ten balls are drawn at random, and you win $1 multiplied by the lowest of the ten numbers. For example, if the balls 2, 8, 14, 15, 27, 45, 54, 62, 89, and 99 were drawn, you would win $2, as 2 is the lowest number.

What is the expected return (i.e. the mean value) of the game?

Quote: charliepatrick

initial ideas/approach

Let the ten first numbers be, from lowest, X Y Z1 Z2...Z8. It transpires you don't care what the Z's actually are. You only need to know 8 numbers between Y and 100 have already been picked.
If you stand you win will X.
If you discard X then there are three possibilities for the next ball B
(i) is less than X, you will win that value and have lost X-B.
(ii) is between X and Y, you will win that value and have won B-X
(iii) is greater than Y, you win will Y and have won Y-X.
As an example 40 50 61 62 63 64 65 66 67 68. (i) 1-39 lose between 1 and 39 (ii) 41-49 gain 1 thru 9 (iii) 51+ gain 10 (note there are 50-8 of these regardless of Z's) So on average you lose 3.61111 by drawing.
In theory this is a function of X and Y.
I'm guessing you could work out the chances of the lowest number being 1,2,3 etc. using permutations. I'm quite so sure how you do the second lowest.

Quote: teliot

Hang on, let me just iterate through the combin(100,10) (approximately 17,310,300,000,000) outcomes.

Show Spoiler

No, really, there's an easy Excel solution. I get 101/11 = $9.18181818 ... as the expected payout. Charge $10 to play the game and you'll make a tidy profit.

Quote: Wizard

Wiz answer

9.181818182

In case anyone is wondering, for the 80/20 case, as in keno, the answer is 3.857143

Quote: teliot

This is pretty much how I envisioned the closed solution being computed as well. But, it's definitely a lot of work and I would want to simulate my answer to verify. So I just ran some quick and dirty C code instead. :)

I: 1 J: 2 Perms: 157366449604 Profit: 1 Win: 2
I: 1 J: 3 Perms: 144520208820 Profit: 1.988888888888889 Win: 2.988888888888889
I: 1 J: 4 Perms: 132601016340 Profit: 2.966666666666667 Win: 3.966666666666667
I: 1 J: 5 Perms: 121550931645 Profit: 3.933333333333333 Win: 4.933333333333334
I: 1 J: 6 Perms: 111315063717 Profit: 4.888888888888889 Win: 5.888888888888889
I: 1 J: 7 Perms: 101841441273 Profit: 5.833333333333333 Win: 6.833333333333333
I: 1 J: 8 Perms: 93080887185 Profit: 6.766666666666667 Win: 7.766666666666667
I: 1 J: 9 Perms: 84986896995 Profit: 7.688888888888889 Win: 8.68888888888889
I: 1 J: 10 Perms: 77515521435 Profit: 8.6 Win: 9.6
I: 1 J: 11 Perms: 70625252863 Profit: 9.5 Win: 10.5 
....
I: 88 J: 89 Perms: 165 Profit: -42.5 Win: 88
I: 88 J: 90 Perms: 45 Profit: -42.477777777777774 Win: 88
I: 88 J: 91 Perms: 9 Profit: -42.46666666666667 Win: 88
I: 88 J: 92 Perms: 1 Profit: -42.46666666666667 Win: 88
I: 89 J: 90 Perms: 45 Profit: -43.48888888888889 Win: 89
I: 89 J: 91 Perms: 9 Profit: -43.477777777777774 Win: 89
I: 89 J: 92 Perms: 1 Profit: -43.477777777777774 Win: 89
I: 90 J: 91 Perms: 9 Profit: -44.48888888888889 Win: 90
I: 90 J: 92 Perms: 1 Profit: -44.48888888888889 Win: 90
I: 91 J: 92 Perms: 1 Profit: -45.5 Win: 91
AvWin: 16.22254798150687 

   totalprofit=0;
   totalperms=0;
   for (i=1; i<=91; i++) {
      for (j=i+1; j<=92; j++) {
         perms=(100-j)*(99-j)*(98-j)*(97-j)*(96-j)*(95-j)*(94-j)*(93-j)/8/7/6/5/4/3/2;
         profit=0;
         if (i>1)   { for (k=1; k<i; k++) {profit+=k-i;};};
         if (j>i+1) { for (k=i+1; k<j; k++) {profit+=k-i;};};
         if (j<92) { for (k=j+9; k<=100; k++) {profit+=j-i;};};
         profit=profit/90;
            outline=" ";
            outline+="I: "+i+" ";
            outline+="J: "+j+" ";
            outline+="Perms: "+perms+" ";
            outline+="Profit: "+profit+" ";
            win=i; if (profit>0) {win=i+profit;};
            outline+="Win: "+win+" ";
            outline+="<BR>";
            document.write(outline);
            totalprofit+=win*perms;
            totalperms+=perms;
      }; }; 
   average=totalprofit/totalperms;
   outline+="AvWin: "+average+" ";
   outline+="<BR>";
   document.write(outline);

Quote: charliepatrick

here's the code which works it out

I haven't checked the answers yet, but this is what it produced.

I: 1 J: 2 Perms: 157366449604 Profit: 1 Win: 2
I: 1 J: 3 Perms: 144520208820 Profit: 1.988888888888889 Win: 2.988888888888889
I: 1 J: 4 Perms: 132601016340 Profit: 2.966666666666667 Win: 3.966666666666667
I: 1 J: 5 Perms: 121550931645 Profit: 3.933333333333333 Win: 4.933333333333334
I: 1 J: 6 Perms: 111315063717 Profit: 4.888888888888889 Win: 5.888888888888889
I: 1 J: 7 Perms: 101841441273 Profit: 5.833333333333333 Win: 6.833333333333333
I: 1 J: 8 Perms: 93080887185 Profit: 6.766666666666667 Win: 7.766666666666667
I: 1 J: 9 Perms: 84986896995 Profit: 7.688888888888889 Win: 8.68888888888889
I: 1 J: 10 Perms: 77515521435 Profit: 8.6 Win: 9.6
I: 1 J: 11 Perms: 70625252863 Profit: 9.5 Win: 10.5 
....
I: 88 J: 89 Perms: 165 Profit: -42.5 Win: 88
I: 88 J: 90 Perms: 45 Profit: -42.477777777777774 Win: 88
I: 88 J: 91 Perms: 9 Profit: -42.46666666666667 Win: 88
I: 88 J: 92 Perms: 1 Profit: -42.46666666666667 Win: 88
I: 89 J: 90 Perms: 45 Profit: -43.48888888888889 Win: 89
I: 89 J: 91 Perms: 9 Profit: -43.477777777777774 Win: 89
I: 89 J: 92 Perms: 1 Profit: -43.477777777777774 Win: 89
I: 90 J: 91 Perms: 9 Profit: -44.48888888888889 Win: 90
I: 90 J: 92 Perms: 1 Profit: -44.48888888888889 Win: 90
I: 91 J: 92 Perms: 1 Profit: -45.5 Win: 91
AvWin: 16.22254798150687 

   totalprofit=0;
   totalperms=0;
   for (i=1; i<=91; i++) {
      for (j=i+1; j<=92; j++) {
         perms=(100-j)*(99-j)*(98-j)*(97-j)*(96-j)*(95-j)*(94-j)*(93-j)/8/7/6/5/4/3/2;
         profit=0;
         if (i>1)   { for (k=1; k<i; k++) {profit+=k-i;};};
         if (j>i+1) { for (k=i+1; k<j; k++) {profit+=k-i;};};
         if (j<92) { for (k=j+9; k<=100; k++) {profit+=j-i;};};
         profit=profit/90;
            outline=" ";
            outline+="I: "+i+" ";
            outline+="J: "+j+" ";
            outline+="Perms: "+perms+" ";
            outline+="Profit: "+profit+" ";
            win=i; if (profit>0) {win=i+profit;};
            outline+="Win: "+win+" ";
            outline+="<BR>";
            document.write(outline);
            totalprofit+=win*perms;
            totalperms+=perms;
      }; }; 
   average=totalprofit/totalperms;
   outline+="AvWin: "+average+" ";
   outline+="<BR>";
   document.write(outline);

         if (i>1)   { for (k=1; k<i; k++) {profit+=k-i;};};
         if (j>i+1) { for (k=i+1; k<j; k++) {profit+=k-i;};};
         if (j<92) { for (k=j+9; k<=100; k++) {profit+=j-i;};};

Quote: teliot

...double check these lines...

Quote: charliepatrick

Show Spoiler

There are 90 possible values for the redrawn ball (I'm assuming it can't be ones you've already got or the one you discarded).
I: 2 J: 3 Perms: 144520208820 Profit: 0.9777777777777777 Win: 2.977777777777778
If the drawn ball is higher than 3 you win 3; only if the new ball drawn in 1 will you not receive 3. The chances of this are 1 in 90 and it loses you 2. So the average is 3 less 2/90. Check.

I: 40 J: 50 Perms: 536878650 Profit: -3.5 Win: 40 T1: -780 T: 39 T1: -735 T: 48 T1: -315 T: 90
From 1 thru 39 you lose 39 thru 1 = -780 (average of -1 and -39 = -20, number of items = 39, total = -780).
From 41 thru 49 you gain 1 thru 9 = +45 (average of 1 and 9 = 5, number of items = 9, total +45.
The remaining 51 to 100 except for the 8 balls already drawn each gain 10. There are 42 os these so total = +420.
420+45-780 = -315. -315/90 = -3.5. Check.

#include <stdio.h>

int main() {
int i, j, k;
double totalprofit = 0;
double totalperms = 0;
double profit, perms, win, num;

   for (i=1; i<=91; i++) {
      for (j=i+1; j<=92; j++) {

         perms=(100.0-j)*(99-j)*(98-j)*(97-j)*(96-j)*(95-j)*(94-j)*(93-j)/(2.0*3*4*5*6*7*8);

         profit=0;
         num = 0;

         if (i>1)   {
           for (k=1; k<i; k++) {
             profit+=(k-i);
             num++;
            }
         }

         if (j>i+1) {
           for (k=i+1; k<j; k++) {
             profit+=(k-i);
             num++;
           }
         }

         if (j<92) {
           for (k=j+9; k<=100; k++) {
             profit+=(j-i);
             num++;
           }
         }

         profit = profit/num;

         win = i+profit;


         totalprofit += (win*perms);
         totalperms += perms;

      }
    }

    printf("%1.6f\n", totalprofit/totalperms);

    return 0;
}

Quote: teliot

optimal strategy for re-draw

Optimal strategy -- thanks to charliepatrick for the code.

Quote: teliot

So, here is a slightly harder version --

A a keno blower has 100 balls, each numbered 1 through 100. Ten balls are drawn at random and recorded. After the ten balls are drawn, the lowest number recorded out of the 10 balls is then discarded and another ball is drawn from the blower to replace it. You then win $1 multiplied by the lowest of the ten numbers.

What is the expected win, to the nearest cent?

Quote: ThatDonGuy

Here's what I have got so far

Suppose the lowest number in the 10 drawn is N.
Let K be the second-lowest number drawn.
The expected value of the lowest number after the draw is (probability that the 11th number < K) x (expected value of the 11th number) + (probability that the 11th number > K) x K.

Of the C(100 - N, 9) sets of 9 numbers > N, the lowest number (i.e. the second-lowest number in the original 10) is:
N + 1, with probability C(99 - N, 8) / C(100 - N, 9)
N + 2, with probability C(98 - N, 8) / C(100 - N, 9)
...
91, with probability C(9, 8) / C(100 - N, 9)
92, with probability C(8, 8) / C(100 - N, 9)

The expected value of the lowest number after the redraw is (K - 1) / 99 x (K + 1) / 2 + (100 - K) / 99 x K
= (K (200 - K) - 1) / 198
where K is the second-lowest number in the original 10

If the original lowest number is 10, the expected value is:
C(89, 8) / C(90, 9) x (11 x 189 - 1) / 198
+ C(88, 8) / C(90, 9) x (12 x 188 - 1) / 198
+ C(87, 8) / C(90, 9) x (13 x 187 - 1) / 198
+ ...
+ C(8, 8) / C(90, 9) x (92 x 8 - 1) / 198

This has to be calculated for each possible lowest number from 1 to 91, then multiply by the probability that it is the original lowest number, and sum the values.

Quote: ThatDonGuy

Here is a math solution (well, sort of) to the "slightly harder" problem with the 100 balls:

Yay Math

First, some explanation of my notation:
C(A,B) is the number of combinations of A things taken B at a time
SUM[B=2..92] {...} is the sum of the value within { } for all B from 2 to 92 inclusive; imagine a capital sigma with "B=2" at the bottom and "92" at the top.

Let A and B be the smallest and second-smallest numbers drawn; note that 1 <= A < B <= 92
There are C(100-B, 8) combinations of the C(100, 10) where this is true
If the 11th number < B, it is the lowest number after A is removed; there are B-2 numbers < B excluding A,
so the expected value of the lowest number in this case is (1 + 2 + ... + (B-1) - A) / (B-2) = ((B-1) B - 2A) / (2 (B-2)).
If the 11th number < B, then B is the lowest number after A is removed.
P(11th number < B) = (B - 2) / 90
P(11th number > B) = (92 - B) / 90
The expected value of the lowest number is
1 / C(100, 10) x SUM[B=2..92] {SUM[A=1..B-1] {C(100-B, 8) x (((B - 2) / 90)((B - 1) B - 2A) / (2 (B - 2)) + B (92 - B) / 90)}}
= 1 / C(100, 10) x SUM[B=2..92] {C(100-B, 8) x SUM[A=1..B-1] {((B - 1) B - 2A) / 180 + 2B (92 - B) / 180}}
= 1 / C(100, 10) x SUM[B=2..92] {C(100-B, 8) x SUM[A=1..B-1] {(183 B - B^2 - 2A) / 180}}
= 1 / (180 C(100, 10)) x SUM[B=2..92] {C(100-B, 8) x (SUM[A=1..B-1] {183 B - B^2} - SUM[A=1..B-1] {2A})}
= 1 / (180 C(100, 10)) x SUM[B=2..92] {C(100-B, 8) x ((B - 1)(183 B - B^2) - 2 (B - 1) B / 2)}
= SUM[B=2..92] {(183 B^2 - B^3 - 182 B) C(100-B, 8)} / (180 C(100, 10))

Okay, I needed a computer to calculate the sum, but it turns out to be 707 / 44.

Quote: teliot

optimal strategy for re-draw

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

Quote: ThatDonGuy

Quote: teliot

optimal strategy for re-draw

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

To be exact...

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.

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?

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.

Quote: charliepatrick

Show Spoiler

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.

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.

Quote: ThatDonGuy

Keep going; you're getting there...

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?

Show Spoiler

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.

working on it got...

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.