# A trigonometry in triangles question?

• Aug 25th 2008, 05:42 AM
fardeen_gen
A trigonometry in triangles question?
In an isoceles triangle, if the altitudes intersect on the inscribed circle, then the cosine of the vertical angle A is?
• Aug 25th 2008, 06:18 AM
fardeen_gen
The options are:
A) 1/9
B) 1/3
C) 2/3
D) None

How do I correlate the facts given in the question?
• Aug 25th 2008, 07:45 AM
Jhevon
Quote:

Originally Posted by fardeen_gen
In an isoceles triangle, if the altitudes intersect on the inscribed circle, then the cosine of the vertical angle A is?

were you given a diagram? because i am not sure where A is
• Aug 25th 2008, 07:54 AM
fardeen_gen
No figure was given. But isn't the location of A obvious?(as it is the vertical angle of the isosceles triangle)
• Aug 25th 2008, 07:56 AM
Jhevon
Quote:

Originally Posted by fardeen_gen
No figure was given. But isn't the location of A obvious?(as it is the vertical angle of the isosceles triangle)

maybe it's just me, but i don't know what "vertical" angle of a triangle means. i'd look it up if i wasn't so lazy :p

EDIT: ok, looked it up. it seems vertical angle has to be in reference to a given side, so i still don't know. are we talking the angle vertical to the side that is (perhaps) of a distinct length? that would make sense, maybe
• Aug 25th 2008, 08:56 AM
fardeen_gen
Yes. The vertical angle is the angle between the two equal lengths.
• Aug 26th 2008, 06:13 AM
fardeen_gen
Using cosine rule does not help. Is the distance between the orthocentre and side BC equal to 2r?
• Aug 28th 2008, 10:43 AM
Moo
Here's a sketch. I think you just have to prove that O, I and A are aligned... looks like it's always the case.

O represents the orthocenter (that is to say the intersection of the altitudes), center of the circumscribed circle c and I the center of the inscribed circle d.

------------------------------------------

Attachment 7623

The inscribed circle was drawn randomly, as well as the position of O on it. I then took a random point on circle c, named A.
Then, I drew the tangents from A to the inscribed circle.
One of this tangents intersects the circumscribed circle in B.
Then I drew the circle e of center A and of radius AB, since ABC is an isosceles triangle.
C must be on this circle e and also on the circumscribed circle c. AC must also be tangent to the inscribed circle d. And C is on these three elements iff A, O and I are aligned.

This doesn't give the solution, but maybe it'll help... ? (Worried)
• Aug 28th 2008, 08:07 PM
fardeen_gen
Thanks for the figure. How is the angle A related to the altitude? I am still not able to figure that out.
• Aug 29th 2008, 05:32 AM
fardeen_gen
My apologies to Jhevon for causing mindless confusion over "vertical angle". The question said "vertex angle" of the isosceles triangle.
• Aug 29th 2008, 06:39 AM
Jhevon
Quote:

Originally Posted by fardeen_gen
My apologies to Jhevon for causing mindless confusion over "vertical angle". The question said "vertex angle" of the isosceles triangle.

eh, don't worry about it. i should have probably known what it was anyway.
• Aug 29th 2008, 06:58 AM
Jhevon
Ok, so geometry isn't my strong suit, but let me submit this anyway. maybe if someone could remind me how does the altitude of an isosceles triangle relate to the other sides. also, i have yet to use the fact that the triangle is inscribed in a circle.

but i found that $\cos A = 2 - \frac 12 \bigg( \frac ab \bigg)^2$ and $\cos A = \frac {2d^2}{b^2} - 1$

here d is the height, b is the side opposite to angle B (one of the other angles)
• Aug 29th 2008, 08:18 PM
fardeen_gen
Is there any theorem connecting altitudes of a triangle and the circles connected to them?