Matematické Fórum

Dacu · 29. 07. 2020 07:51

Hello all,

Between the angles of a triangle there is the relation $A^2 =B^2+C^2$ .Find all triangles for which $A$ , $B$ , $C$ are sexagesimal angles and $A , B , C\in \mathbb N^*$ .

All the best,

Dacu

kerajs · 29. 07. 2020 16:08

1) A=82 , B=80 , C=18
2) A=82 , B=18 , C=80
3) A=78 , B=72 , C=30
4) A=78 , B=30 , C=72
5) A=75 , B=60 , C=45
6) A=75 , B=45 , C=60

edison · 29. 07. 2020 19:25

Možná je to hloupá otázka, ale co znamená "sexagesimal"?

kerajs · 29. 07. 2020 19:42

https://cs.wikipedia.org/wiki/%C5%A0ede … 1_soustava

Dacu · 30. 07. 2020 05:45 — Editoval Dacu (30. 07. 2020 05:47)

kerajs napsal(a):
1) A=82 , B=80 , C=18
2) A=82 , B=18 , C=80
3) A=78 , B=72 , C=30
4) A=78 , B=30 , C=72
5) A=75 , B=60 , C=45
6) A=75 , B=45 , C=60

Hello,

Are there other triangles? If not , then prove that these are the only triangles?Thank you very much!

All the best,

Dacu

kerajs · 30. 07. 2020 08:20

$& A^2=B^2+C^2 \\ & (180-B-C)^2=B^2+C^2 \\ & (180-B)(180-C)=180\cdot 90\\ & 180\cdot 90=2^33^45^2=100\cdot 162=108\cdot150=120\cdot 135$

Dacu · 30. 07. 2020 14:43

kerajs napsal(a):
$& A^2=B^2+C^2 \\ & (180-B-C)^2=B^2+C^2 \\ & (180-B)(180-C)=180\cdot 90\\ & 180\cdot 90=2^33^45^2=100\cdot 162=108\cdot150=120\cdot 135$

Hello,

I do not understand!Where do we get $(180-B)(180-C)=180\cdot 90$ ?Thank you very much!

All the best,

Dacu

misaH · 30. 07. 2020 16:02 — Editoval misaH (30. 07. 2020 16:03)

↑ Dacu:

It's not so difficult!
(When I can do it...)

$& A^2=B^2+C^2 \\ & (180-B-C)^2=B^2+C^2 \\ & (180-B)(180-C)=180\cdot 90\\ & 180\cdot 90=2^33^45^2=100\cdot 162=108\cdot150=120\cdot 135$

$& A^2=B^2+C^2 \\ & (180-B-C)^2=B^2+C^2 \\ &180^2-2\cdot180(B+C)+(B+C)^2=B^2+C^2$
So
$\color{blue}BC=180(B+C)-90\cdot 180$
$(180-B)(180-C)=180^2-180(B+C)+\color{blue}BC$
$& (180-B)(180-C)=180\cdot 90$

Dacu · 30. 07. 2020 16:33

misaH napsal(a):
↑ Dacu:

It's not so difficult!
(When I can do it...)

$& A^2=B^2+C^2 \\ & (180-B-C)^2=B^2+C^2 \\ & (180-B)(180-C)=180\cdot 90\\ & 180\cdot 90=2^33^45^2=100\cdot 162=108\cdot150=120\cdot 135$

$& A^2=B^2+C^2 \\ & (180-B-C)^2=B^2+C^2 \\ &180^2-2\cdot180(B+C)+(B+C)^2=B^2+C^2$
So
$\color{blue}BC=180(B+C)-90\cdot 180$
$(180-B)(180-C)=180^2-180(B+C)+\color{blue}BC$
$& (180-B)(180-C)=180\cdot 90$

Hello,

Correct!I understand!How do we prove that there are only 6 triangles?Thank you very much!

All the best,

Dacu

kerajs · 30. 07. 2020 17:09

Really?
$& \begin{cases} 180-B=100 \\ 180-C=162 \end{cases} \vee \begin{cases} 180-B=162 \\ 180-C=100 \end{cases} \vee \begin{cases} 180-B=108 \\ 180-C=150 \end{cases} \vee \begin{cases} 180-B=150 \\ 180-C=108 \end{cases} \vee \\ & \vee \begin{cases} 180-B=120 \\ 180-C=135 \end{cases} \vee \begin{cases} 180-B=135 \\ 180-C=120 \end{cases}$

Dacu · 31. 07. 2020 06:49

kerajs napsal(a):
Really?
$& \begin{cases} 180-B=100 \\ 180-C=162 \end{cases} \vee \begin{cases} 180-B=162 \\ 180-C=100 \end{cases} \vee \begin{cases} 180-B=108 \\ 180-C=150 \end{cases} \vee \begin{cases} 180-B=150 \\ 180-C=108 \end{cases} \vee \\ & \vee \begin{cases} 180-B=120 \\ 180-C=135 \end{cases} \vee \begin{cases} 180-B=135 \\ 180-C=120 \end{cases}$

Hello,

Yes!Why can't there be other combinations?Thank you very much!

All the best,

Dacu

kerajs · 31. 07. 2020 07:56

Because:
$& \begin{cases} 0<180-B<180 \\ 0<180-C<180 \\ (180-B)(180-C)=180\cdot 90 \end{cases} \Rightarrow \begin{cases} 90<180-B<180 \\ 90<180-C<180 \\ (180-B)(180-C)=2^33^45^2 \end{cases}$

Dacu · 05. 08. 2020 08:18 — Editoval Dacu (05. 08. 2020 08:28)

kerajs napsal(a):
Because:
$& \begin{cases} 0<180-B<180 \\ 0<180-C<180 \\ (180-B)(180-C)=180\cdot 90 \end{cases} \Rightarrow \begin{cases} 90<180-B<180 \\ 90<180-C<180 \\ (180-B)(180-C)=2^33^45^2 \end{cases}$

Hello,

My reasoning:

1) $A^2=B^2+C^2$
2) $(180-A)^2=(B+C)^2$
3) $B\cdot C=180\cdot(90-A)$ , $A<90^\circ$
4) $B+C=180-A$ , $B\cdot C=180\cdot (90-A)$ , $x^2-(180-A)\cdot x+180\cdot 90-180\cdot A=0$ , $x_1=B$ , $x_2=C$ , $B,C=\frac{(180-A)\mp \sqrt{A^2+360\cdot A-180^2}}{2}$
5) $A^2+360\cdot A-180^2>0$ , $75\leq A<90$
6) $A^2+360\cdot A-180^2=u^2$ , $u\in \mathbb N^*$
7) $A^2-u^2=360\cdot (90-A)$ , $(A-u)(A+u)=360\cdot (90-A)$ , $u<A<90$
8) $A=75^\circ$ , $A=78^\circ$ , $A=82^\circ$

All the best,

Dacu