# Putnam Problem of the Day (3)

• Jun 10th 2010, 10:37 PM
simplependulum
Putnam Problem of the Day (3)
This problem was really challenging to me , I've spent much time on this but maybe it is easy to you guys . Enjoy !

Triangle $ABC$ has the following property : there is an interior point P such that $\angle PAB = 10^o$ , $\angle PBA = 20^o$ , $\angle PCA = 30^o$ and $\angle PAC = 40^o$ . Prove that triangle $ABC$ is isosceles .

Moderator edit: Approved Challenge question.
• Jun 11th 2010, 01:13 AM
undefined
Quote:

Originally Posted by simplependulum
This problem was really challenging to me , I've spent much time on this but maybe it is easy to you guys . Enjoy !

Triangle $ABC$ has the following property : there is an interior point P such that $\angle PAB = 10^o$ , $\angle PBA = 20^o$ , $\angle PCA = 30^o$ and $\angle PAC = 40^o$ . Prove that triangle $ABC$ is isosceles .

Moderator edit: Approved Challenge question.

Maybe I'm close? I don't know how to show without a calculator that

Spoiler:

$\sin80^{\circ}=4\cdot\sin20^{\circ}\cdot\sin50^{\c irc}\cdot\sin70^{\circ}$

which I got through using law of sines and then guessing which sides are congruent (not very elegant I know).
• Jun 11th 2010, 02:00 AM
TheCoffeeMachine
Quote:

Originally Posted by undefined
Maybe I'm close? I don't know how to show without a calculator that

Spoiler:

$\sin80^{\circ}=4\cdot\sin20^{\circ}\cdot\sin50^{\c irc}\cdot\sin70^{\circ}$

which I got through using law of sines and then guessing which sides are congruent (not very elegant I know).

Morrie's law?
• Jun 11th 2010, 02:13 AM
undefined
Quote:

Originally Posted by TheCoffeeMachine

Yes that works perfectly! Thanks.
• Jun 11th 2010, 02:14 AM
simplependulum
$\sin80^{\circ}=4\cdot\sin20^{\circ}\cdot\sin50^{\c irc}\cdot\sin70^{\circ}$

Of course it is correct , by writing $\sin{70^o} = \cos{20^o}~ ~,~ \sin{50^o} = \cos{40^o}$

$4\cdot\sin20^{\circ}\cdot\sin50^{\circ}\cdot\sin70 ^{\circ} = (2\sin{20^o}\cos{20^o}) (2\cos{40^o})$

$= 2\sin{40^o}\cos{40^o} = \sin{80^o}$

It is a nice trigonometric solution to the problem , mine is a pure geometrical solution , see if you can find any (Happy) .
• Jun 11th 2010, 02:25 AM
undefined
Quote:

Originally Posted by simplependulum
$\sin80^{\circ}=4\cdot\sin20^{\circ}\cdot\sin50^{\c irc}\cdot\sin70^{\circ}$

Of course it is correct , by writing $\sin{70^o} = \cos{20^o}~ ~,~ \sin{50^o} = \cos{40^o}$

$4\cdot\sin20^{\circ}\cdot\sin50^{\circ}\cdot\sin70 ^{\circ} = (2\sin{20^o}\cos{20^o}) (2\cos{\color{red}50\color{black}^o})$

$= 2\sin{40^o}\cos{40^o} = \sin{80^o}$

It is a nice trigonometric solution to the problem , mine is a pure geometrical solution , see if you can find any (Happy) .

Can't believe I've forgotten the double angle formulas already. (Surprised) Only somewhat justifiable by the fact that I haven't used them in about 5 years.

Not sure I'll have any luck with that pure geometric solution though.

By the way there was a small typo in your post, in red above.
• Jun 11th 2010, 02:31 AM
simplependulum
Oh it is a typo , thanks for the correction , it is clear now (Happy) .