Let A B C be a triangle and h a be the altitude through A . Prove that b + c 2 ≥ a 2 + 4 h a 2 (As usual a , b , c denote the sides B C , C A , A B respectively)

Draw a line l parallel to B C through A and reflect A C in this line to get A D . Let C D intersect l in P . Join B D . Observe that C P = P D = A Q = h a , A Q being the altitude through A . Now, b + c = A C + A B = A D + A B ≥ B D = C D 2 + C B 2 = 2 h a 2 + a 2 = 4 h a 2 + a 2 (by pythagoras theorem) Equality occurs if and only if B , A , D are collinear, i.e., if and only if A D = A B (as A P is parallel to B C and bisects D C ) and this is equivalent to A C = B C .

Let A B C be a triangle and h a be the altitude through A . Prove that b + c 2 ≥ a 2 + 4 h a 2 (As usual a , b , c denote the sides B C , C A , A B respectively)

Draw a line l parallel to B C through A and reflect A C in this line to get A D . Let C D intersect l in P . Join B D . Observe that C P = P D = A Q = h a , A Q being the altitude through A . Now, b + c = A C + A B = A D + A B ≥ B D = C D 2 + C B 2 = 2 h a 2 + a 2 = 4 h a 2 + a 2 (by pythagoras theorem) Equality occurs if and only if B , A , D are collinear, i.e., if and only if A D = A B (as A P is parallel to B C and bisects D C ) and this is equivalent to A C = B C .

Suppose A B C D is a cyclic quadrilateral inscribed in a circle of radius one unit. If A B · B C · C D · D A ≥ 4 , then prove that A B C D is a square.

Let A B = a , B C = b , C D = c , D A = d . We are given that a b c d ≥ 4 . Using Ptolemy's theorem and the fact that each diagonal cannot exceed the diameter of the circle, we get a c + b d = A C · B D ≤ 4 . . . . . . . i By AM - GM inequality for two terms a c & b d , we get a c + b d ≥ 2 a b c d = 2 4 = 4 . . . . . . . ii From both the equations, we conclude that a c + b d = 4 . This forces A C · B D = 4 giving A C = B D = 2 . Each of A C and B D is thus a diameter. This implies that A B C D is a rectangle. Note : a c - b d 2 = a c + b d 2 - 4 a b c d ≤ 16 - 16 = 0 and hence a c = b d = 2 . We know that opposite sides of a rectangle are equal. Thus, we get a = c = a c = 2 and similarly b = d = 2 . It now follows that A B C D is a square.

E M B I B E

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 73

EASY

Olympiad

IMPORTANT

Earn 100

Let $A B C$ be a triangle and $h_{a}$ be the altitude through $A$ . Prove that ${(b + c)}^{2} \geq a^{2} + 4 h_{a}^{2}$ (As usual $a, b, c$ denote the sides $B C, C A, A B$ respectively)

Important Questions on Geometry

HARD

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 74

Let

P

be an interior point of a triangle

A B C

and let

B P

and

C P

meet

A C

and

A B

in

E

and

F

respectively. If

[B P F] = 4, [B P C] = 8

and

[C P E] = 13

, find

[A F P E]

. (Here,

[]

denotes the area of a triangle or a quadrilateral as the case may be)

MEDIUM

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 75

Suppose $A B C D$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $A B \cdot B C \cdot C D \cdot D A \geq 4$ , then prove that $A B C D$ is a square.