MHB Determine the ratio of two angles in a triangle.

AI Thread Summary
In triangle ABC, the relationship between the sides is given by the equation \(\frac{BC}{AB-BC}=\frac{AB+BC}{AC}\), leading to the conclusion that the ratio of angles \(\frac{\angle A}{\angle C}\) can be determined. By applying the cosine rule and manipulating the equations, it is shown that \(\angle C = 2\angle A\). The discussion highlights the geometric interpretation of the problem, particularly focusing on isosceles triangles formed during the analysis. Participants express appreciation for the clarity and effectiveness of the methods used to solve the problem. Ultimately, the findings emphasize the connection between angle ratios and side lengths in triangle geometry.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $$ABC$$ be a triangle such that $$\frac{BC}{AB-BC}=\frac{AB+BC}{AC}$$.

Determine the ratio $$\frac{\angle A}{\angle C}$$.
 
Mathematics news on Phys.org
anemone said:
Let $$ABC$$ be a triangle such that $$\frac{BC}{AB-BC}=\frac{AB+BC}{AC}$$.

Determine the ratio $$\frac{\angle A}{\angle C}$$.
Write $a,\ b,\ c$ for the sides opposite the angles $A,\ B,\ C$ respectively. You are told that $\dfrac a{c-a} = \dfrac{c+a}b$, so that $ab = c^2-a^2$. The cosine rule tells you that $c^2 = a^2+b^2-2ab\cos C$. Combining those equations, you get $ab = b^2 -2ab\cos C$, so that $a = b-2a\cos C$. Now draw a picture.

angles.png

The base of the isosceles triangle $BCD$ is twice the projection of $BC$ onto the side $CA$, namely $2a\cos C$. Therefore $DA = b-2a\cos C = a$. Thus $BDA$ is also isosceles, and it follows that $\angle C = \angle BDC = 2\angle A$.
 
Last edited:
Opalg said:
Write $a,\ b,\ c$ for the sides opposite the angles $A,\ B,\ C$ respectively. You are told that $\dfrac a{c-a} = \dfrac{c+a}b$, so that $ab = c^2-a^2$. The cosine rule tells you that $c^2 = a^2+b^2-2ab\cos C$. Combining those equations, you get $ab = b^2 -2ab\cos C$, so that $a = b-2a\cos C$. Now draw a picture.

angles.png

The base of the isosceles triangle $BCD$ is twice the projection of $BC$ onto the side $CA$, namely $2a\cos C$. Therefore $DA = b-2a\cos C = a$. Thus $BDA$ is also isosceles, and it follows that $\angle C = \angle BDC = 2\angle A$.

Awesome! :cool: I used a totally different method and yours is without a doubt, so much better than mine!

Thanks for showing me that we could actually solve a geometric problem using such a method and to be honest with you, I always love to read how you would approach a problem...:o

Thank you so much, Opalg!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top