Maths Olympiad Prep

Library / /10 of 22

Geometry Difficulty 6.2 National olympiad Prove it South Africa

In obtuse triangle ABCABC, with the obtuse angle at AA, let DD, EE, FF be the feet of the altitudes through AA, BB, CC respectively. DEDE is parallel to CFCF, and DFDF is parallel to the angle bisector of ∠BAC\angle BAC. Find the angles of the triangle.

Solution

Figure 1

We denote the angles of the triangle by α=∠BAC\alpha = \angle BAC, β=∠ABC\beta = \angle ABC and γ=∠ACB\gamma = \angle ACB. XX is the intersection of BCBC with the angle bisector of ∠BAC\angle BAC. Since ∠BDA=∠BEA=90∘\angle BDA = \angle BEA = 90^\circ, both DD and EE lie on the circle with diameter ABAB. Thus AEBDAEBD is cyclic, which implies that
∠EDB=∠EAB=180∘−α. \angle EDB = \angle EAB = 180^\circ - \alpha.
It is given that DEDE and CFCF are parallel, hence
∠EDB=∠FCB=90∘−∠FBC=90∘−β, \angle EDB = \angle FCB = 90^\circ - \angle FBC = 90^\circ - \beta,
so β=α−90∘\beta = \alpha - 90^\circ. Likewise,
∠FDC=∠FAC=180∘−α, \angle FDC = \angle FAC = 180^\circ - \alpha,
and
∠FDC=∠AXC=180∘−∠ACX−∠XAC=180∘−γ−α/2, \angle FDC = \angle AXC = 180^\circ - \angle ACX - \angle XAC = 180^\circ - \gamma - \alpha/2,
so γ=α/2\gamma = \alpha/2. Since α+β+γ=180∘\alpha + \beta + \gamma = 180^\circ, this gives us
α+(α−90∘)+α/2=180∘, \alpha + (\alpha - 90^\circ) + \alpha/2 = 180^\circ,
and thus α=108∘\alpha = 108^\circ, β=18∘\beta = 18^\circ and γ=54∘\gamma = 54^\circ.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.