Schaum easy outlines-The Math Problems Notebook

Valentin Boju
Louis Funar
The Math Problems Notebook
Birkh¨auser
Boston • Basel • Berlin


7
Geometry Solutions
7.1 Synthetic Geometry
Problem 3.1. Let I be the center of the circle inscribed in the triangle ABC and
consider the points α, β, γ situated on the perpendiculars from I on the sides of the
triangle ABC such that
Iα = Iβ = Iγ .
Prove that the lines Aα, Bβ,Cγ are concurrent.
Solution 3.1. The lines Aα, Bβ,Cγ are concurrent if and only if they satisfy Ceva’s
theorem. Let us draw A1A2  BC, where A1 ∈ AB, A2 ∈ AC, and α ∈ A1A2.
We also draw their analogues. Using Thales’ theorem, the Ceva condition is reduced
to
&
cyclic
αA2
αA1 = 1.

Now I is situated on all three bisectors and Iγ = Iα. By symmetry we have
C1γ = A2α and their analogues. This shows that the Ceva condition is satisfied.

Problem 3.2. We consider the anglexOy and a pointA ∈ Ox. Let (C) be an arbitrary
circle that is tangent to Ox and Oy at the points H and D, respectively. Set AE for
the tangent line drawn from A to the circle (C) that is different from AH. Show that
the line DE passes through a fixed point that is independent of the circle (C) chosen
above.
Solution 3.2. Let A be the point on Oy such that OA = OA. Let us further
consider the intersection points AE ∩ Oy = {F} and DE ∩ AA = {P}.

We use Menelaus’s theorem in the triangle AFA with the line DEP as transversal.
We have
|PA|
|PA| · |EA|
|EF| · |DF|
|DA| = 1.
But one knows that |EA| = |AH|, |EF| = |DF|, and |AD| = |AH|. Then |PA|
|PA| =
1, and therefore P is the midpoint of |AA|. Consequently, P is a fixed point.