National Junior 2009

[sourav das]

A triangle is a polygon with three sides and a strictly positive area. One
angle of a triangle is twice of another angle of the same triangle. An angle of
this triangle is $120$ degree. The bisector of the second largest triangle
intersects its opposite side at point D. The distance of D from the vertex
containing the largest angle is $10 cm$. If the length of the largest side of this
triangle is $2x$, then a relationship like the following is true:
$x^4-C_1x^3-C_2x^2-C_3x+1875=0$
Find the value of $C_1$, $C_2$ and $C_3$ analytically.

[Honestly speaking, for me, it is the hardest problem in every section of BDMO of all the time. So give your best shot....]

[Nadim Ul Abrar]

Will It be enough to find a triple of $(C_1,C_2,C_3)$ that satisfy the relation ??

[nafistiham]

Just gotta use cos rule and similar triangles
But,too much calculation to do at a stretch.

[abid85]

will you give the full solution pls??

[tanmoy]

Sourav da,have you solved the problem!If you,please give the solution.

[Nayeemul Islam Swad]

I've not got any solution yet after trying to solve this problem for 2 years. If anyone got the solution please show it

[asif e elahi]

I think there is something wrong in this problem.We can find out x without solving this quartic equation.

[*Mahi*]

The question asks to find out the values of $C_1, C_2, C_3$.

[asif e elahi]

I found $x=5\sqrt{3}\cot 20^{\circ}$.If we write $x^{4}-c_{1}x^{3}-c_{2}x^{2}-c_{3}x+1875=(x-5\sqrt{3}cot 20^{\circ})(x-a)(x-b)(x-c)$,where $5\sqrt{3}cot 20^{\circ}abc=1875$,we get infinite a,b,c.So we get infinite $c_{1},c_{2},c_{3}$.

[samiul_samin]

I also agree that it is the hardest of junior catagory.I don't know how to solve the last equation.