pythagorean thm: triangle's sides have lengths xx, yy, sqrt[xxyy]; find x+y.

[kastamonu]

triangle:

A*
 |  `   sqrt[xxyy]
 |yy    `
 |          `
 |     xx       `
B*---------------*C

ABC is a right triangle
AB is perpendicular to BC
|AB| is yy units
|BC| is xx units
|AC| is sqrt[xxyy] units
xx and yy are two digits
xxyy is a four-digit natural number.
Find the value of x + y.

(10y + y)² + (10x + x)² = xxyy

121y² + 121x² = 1100x + 11y

11y² + 11x² = 100x + y

< LINK REMOVED >

I got the answer as 8833
8+3=11
Is there a shorter way?

 

[pka]

https://postimg.org/image/afo2xvwdl/
I got the answer as 8833
8+3=11 Is there a shorter way?

I have no idea because yours is the worst notation I think I have ever seen.
Lets try to clean it up. Let each of \(\displaystyle \{x_1,~x_2,~y_1,~y_2\} \) be a digit.
Now what you label \(\displaystyle xx=10x_2+x_1~\&~yy=10y_2+y_1 \)
Thus \(\displaystyle \large (xx)(yy)=(10x_2)(10y_2)+(10x_2)(y_1)+(x_1)(10y_2)+(x_1)(y_1) \)

If that is not a correct reading, I find the question meaningless.

 

[Ishuda]

I have no idea because yours is the worst notation I think I have ever seen.
Lets try to clean it up. Let each of \(\displaystyle \{x_1,~x_2,~y_1,~y_2\} \) be a digit.
Now what you label \(\displaystyle xx=10x_2+x_1~\&~yy=10y_2+y_1 \)
Thus \(\displaystyle \large (xx)(yy)=(10x_2)(10y_2)+(10x_2)(y_1)+(x_1)(10y_2)+(x_1)(y_1) \)

If that is not a correct reading, I find the question meaningless.

The way I read it is that x₁, x₂, y₁, and y₂ are single digits, all between 1 and 9\(\displaystyle ^*\) with x₁=x₂=x and y₁=y₂=y. Thus
xx = 10 x + x = 11 x
yy = 10 y + y = 11 y
xxyy = 1000 x + 100 x + 10 y + y
so the Pythagorean Theorem gives
11 x² + 11 y² = 100 x + y
<sub>This can be reformatted as the equation for a circle which might give further in site into solutions (if any exist).

*After returning from the corner edited to say 'all between 0 and 10' is the correct statement as so nicely pointed out by pka. Thank you.</sub>

 

[stapel]

I got the answer as 8833
8+3=11
Is there a shorter way?

Since you haven't shown your steps, between the quadratic equation and your final answer, there is no way to know "how long" your solution went, so there is no way to know if there might be a "shorter way". The following, being the second reply you received, may be a shorter way, though that poster ends his working at the same point.

...Thus
xx = 10 x + x = 11 x
yy = 10 y + y = 11 y
xxyy = 1000 x + 100 x + 10 y + y
so the Pythagorean Theorem gives
11 x² + 11 y² = 100 x + y
This can be reformatted as the equation for a circle which might give further in site into solutions (if any exist).

Since each of you ended up with one equation in two variables, there can be no strictly algebraic solution. One must, at this point, resort to guess-n-check, or else perhaps apply the methods of Dipphantine equations...?

The circle equation comes out to be:

. . . . .\(\displaystyle \left(x\, -\, \dfrac{50}{11}\right)^2\, +\, \left(y\, -\, \dfrac{1}{22}\right)^2\, =\, \left(\dfrac{\sqrt{\strut 10,001\,}}{22}\right)^2\)

I don't think this helps, particularly....

 

[Ishuda]

Since you haven't shown your steps, between the quadratic equation and your final answer, there is no way to know "how long" your solution went, so there is no way to know if there might be a "shorter way". The following, being the second reply you received, may be a shorter way, though that poster ends his working at the same point.


Since each of you ended up with one equation in two variables, there can be no strictly algebraic solution. One must, at this point, resort to guess-n-check, or else perhaps apply the methods of Dipphantine equations...?

The circle equation comes out to be:

. . . . .\(\displaystyle \left(x\, -\, \dfrac{50}{11}\right)^2\, +\, \left(y\, -\, \dfrac{1}{22}\right)^2\, =\, \left(\dfrac{\sqrt{\strut 10,001\,}}{22}\right)^2\)

I don't think this helps, particularly....

It gives you limits on the possible solutions since they must lie on the circle.

 

[pka]

The way I read it is that x₁, x₂, y₁, and y₂ are single digits, all between 1 and 9 with x₁=x₂=x and y₁=y₂=y.

In writing mathematics, to say that x is an integer between 1 and 9 means \(\displaystyle 2\le x\le 8\).
If you want \(\displaystyle 1\le x\le 9\) then write it as 'x is an integer from 1 to 9'.

Hilbert call attention to this distinction with his betweeness axioms for geometry.

 

[hoosie]

ABC is a right triangle
AB is perpendicular to BC
|AB| is yy units
|BC| is xx units
|AC| is sqrt[xxyy] units
xx and yy are two digits
xxyy is a four-digit natural number.
Find the value of x + y

By Theorem of Pythagoras (
For each 4 digit number below we need to:
* square second and fourth digits
* add the squares and write down final digit of answer
* determine whether the final digit you have recorded is the
same as the last digit of the 4 digit number

1122 1133 1144 1155 1166 1177 1188 1199
2233 2244 2255 2266 2277 2288 2299
3344 3355 3366 3377 3388 3399
4455 4466 4477 4488 4499
5566 5577 5588 5599
6677 6688 6699
7788 7799
8899
2211 3311 4411 5511 6611 7711 8811 9911
3322 4422 5522 6622 7722 8822 9922
4433 5533 6633 7733 8833 9933
5544 6644 7744 8844 9944
6655 7755 8855 9955
7766 8866 9966
8877 9977
9988

For example consider 2255
- square second and fourth digits: 2^2 = 4 5^2 = 25
- add the squares and write down last digit of answer : 4 + 25 = 29 Last digit = 9
- determine if the last digit you have recorded is the same as the last digit of 2255
9 is not the same as 5 so cross 2255 off the list
Repeat for the other 4 digit numbers:

Now consider 8833
See if you can fill in the boxes below:
- square second and fourth digits: 8^2 = 3^2 = ?
- add the squares and write down last digit of answer : + ?= ️ Last digit = ?
- determine if the last digit recorded is the same as the last digit of ????

?is the same as the last digit of ????
Remember we need to show (xx)^2 + (yy) ^2 = xxyy
Does ??x? ?+ ??x?? = ????? If it does you are correct!

Therefore xxyy = ????
where x = ?, y = ?
It follows x + y = ?

 

[Ishuda]

Ishuda, to the corner for 19 minutes...

On my way!

 

[Ishuda]

Interesting ?

Making it (aa)^2 + (bb)^2 = (cc)^2 ****
where a,b,c are digits>0

(10a + a)^2 + (10b + b)^2 = (10c + c)^2
Simplifies to : a^2 + b^2 = c^2

Take a=3. b=4, c=5: 33^2 + 44^2 = 55^2

And that's only case where that happens...

Btw, I think the OP is incorrectly posted...should be ****

Other than the trivial solution. Of course, if you are one of those who do not include 0 as a digit, please forget I 'said' anything.

 

[pka]

Making it (aa)^2 + (bb)^2 = (cc)^2 ****
where a,b,c are digits>0
(10a + a)^2 + (10b + b)^2 = (10c + c)^2
Simplifies to : a^2 + b^2 = c^2
Take a=3. b=4, c=5: 33^2 + 44^2 = 55^2
And that's only case where that happens...

Thank you Dennis, I agree with your solution.

In reply #2 to the original poster, I said

yours is the worst notation I think I have ever seen.

Who would use \(\displaystyle \sqrt{xxyy}\text{ for }\sqrt{10^3x+10^2x+10y+y}~?\)

 

[Deleted member 4993]

triangle:

A*
 |  `   sqrt[xxyy]
 |yy    `
 |          `
 |     xx       `
B*---------------*C

ABC is a right triangle
AB is perpendicular to BC
|AB| is yy units
|BC| is xx units
|AC| is sqrt[xxyy] units
xx and yy are two digits
xxyy is a four-digit natural number.
Find the value of x + y.

(10y + y)² + (10x + x)² = xxyy

121y² + 121x² = 1100x + 11y

11y² + 11x² = 100x + y → 11(x²+y²) = x0y → (x0y is divisible by 11) → (x+y) - 0 = 11*m → (x+y) is divisible by 11 → x+y = 11

I got the answer as 8833
8+3=11
Is there a shorter way?

.

 

[pka]

Giddyup...giddyup...I'm aridin' in to the defense of poor "0"
05^2 + 12^2 = 13^2
0505^2 + 1212^2 = 1313^2
050505^2 + 121212^2 = 131313^2
Thheeerrrreeeeee: 0 feels useful for a change!

Thank again Denis. Of course zero is a digit.

​

 

[Deleted member 4993]

Giddyup...giddyup...I'm aridin' in to the defense of poor "0"

05^2 + 12^2 = 13^2

0505^2 + 1212^2 = 1313^2

050505^2 + 121212^2 = 131313^2

Thheeerrrreeeeee: 0 feels useful for a change!

I think - it is 11 that is doing the trick riding.....