Triangle Problem | PRMO-2018 | Problem No-24

Pre College Mathematics

Prmo-2018, Problem-24

\(27\)

Given that $\mathrm{N}$ is the number of triangles of different shapes. Therefore the different shapes of triangle the angles will be change . at first we have to find out the posssible orders of the angles that the shape of the triangle will be different...

Now can you finish the problem?

case 1 : when $ x \geq 1$ & $y \geq 3 \geq 1$
$$
x+y+z=180
$$
$={ }^{179} \mathrm{C}_{2}=15931$
Case 2 : When two angles are same
$$
2 x+y=180
$$
1,1,178
2,2,176
$\vdots$
89,89,2

But we have one case $60^{\circ}, 60^{\circ}, 60^{\circ}$
$$
\text { Total }=89-1=88
$$
Such type of triangle $=3(88)$
When 3 angles are same $=1(60,60,60)$
So all distinct angles's triangles
$$
\begin{array}{l}
=15931-(3 \times 88)-1 \
\neq 3 ! \
=2611
\end{array}
$$
Now, distinct triangle $=2611+88+1$
$
=2700 \
N=2700 \
\frac{N}{100}=27 \
$

Pre College Mathematics

Prmo-2018, Problem-24

\(27\)

Given that $\mathrm{N}$ is the number of triangles of different shapes. Therefore the different shapes of triangle the angles will be change . at first we have to find out the posssible orders of the angles that the shape of the triangle will be different...

Now can you finish the problem?

case 1 : when $ x \geq 1$ & $y \geq 3 \geq 1$
$$
x+y+z=180
$$
$={ }^{179} \mathrm{C}_{2}=15931$
Case 2 : When two angles are same
$$
2 x+y=180
$$
1,1,178
2,2,176
$\vdots$
89,89,2

But we have one case $60^{\circ}, 60^{\circ}, 60^{\circ}$
$$
\text { Total }=89-1=88
$$
Such type of triangle $=3(88)$
When 3 angles are same $=1(60,60,60)$
So all distinct angles's triangles
$$
\begin{array}{l}
=15931-(3 \times 88)-1 \
\neq 3 ! \
=2611
\end{array}
$$
Now, distinct triangle $=2611+88+1$
$
=2700 \
N=2700 \
\frac{N}{100}=27 \
$