1. Let ABC be a right-angled triangle with ∠B = 90◦. Let I be the incentre of ABC. Let AI extended intersect BC in F. Draw a line perpendicular to AI at I. Let it intersect AC in E. Prove that
IE = IF.
please give me full exlantaion and write it answer only
Answers
Solution: Extend EI to meet CB extended in D.
First observe that ADBI is a cyclic quadrilateral
since ∠AID = ∠ABD. Hence ∠ADI = ∠ABI =
45◦
. Hence ∠DAI = 45◦
. Therefore IA = ID.
Consider the triangles AIE and DIF. Both are
right triangles. Moreover ∠IAE = ∠IAB = ∠IDB.
Since IA = ID, the triangles are congruent. This
means IE = IF.
2. Let a, b, c be positive real numbers such that
a
1 + b
+
b
1 + c
+
c
1 + a
= 1.
Prove that abc ≤ 1/8.
Solution: This is equivalent to
Xa(1 + c)(1 + a) = (1 + a)(1 + b)(1 + c).
This simplifies to
Xa
2 +
Xa
2
c = 1 + abc
Using AM-GM inequality, we have
1 + abc =
Xa
2 +
Xa
2
c ≥ 3(abc)
2/3 + 3abc.
Let x = (abc)
1/3
. Then
3x
2 + 2x
3 ≤ 1.
This can be written as (x + 1)2
(2x − 1) ≤ 0. Hence x ≤ 1/2. Thus
abc ≤
1
8
.