A triangle A B C has incentre I . Points X Y are located on the line segments A B , A C respectively so that B X · A B = I B 2 and C Y · A C = I C 2 . Given that X , I , Y are collinear, find the possible values of the measure of angle A .

From the relation B I 2 = B X · B A We see that B I is a tangent to the circle passing through A , X , I at I . Hence ∠ B I X = ∠ B A I = A 2 . . . 1 [Alternatively, one observes that in triangles B I X and B A I , ∠ I B X is common and B I / B X = B A / B I . Consequently, the two triangles are similar, implying 1 .] Similarly, from the relation C I 2 = C Y · C A , We obtain ∠ C I Y = ∠ C A I = A 2 . . . 2 It is known that ∠ B I C = 90 ° + A 2 . . . 3 From (1), (2), (3) and the fact that X , I , Y are collinear, we obtain A 2 + A 2 + 90 ° + A 2 = 180 ° Solving we get A = 60 ° .

E M B I B E

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 55

MEDIUM

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IMPORTANT

Earn 100

Suppose $A B C D$ is a cyclic quadrilateral and $x, y, z$ are the distances of $A$ from the lines $B D, B C, C D$ respectively. Prove the

$\frac{B D}{x} = \frac{B C}{y} + \frac{C D}{z} .$

Important Questions on Geometry

MEDIUM

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IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 56

Suppose

A B C D

is a convex quadrilateral and

P Q

are the midpoints of

C D, A B

. Let

A P, D Q

meet in

X

and

B P, C Q

meet in

Y

. Prove that

[A D X] + [B C Y] = [P X Q Y]

. How does the conclusion alter if

A B C D

is not a convex quadrilateral?

HARD

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 57

Suppose $P$ is an interior point of a triangle $A B C$ and $A P, B P, C P$ meet the opposite sides $B C, C A, A B$ in $D, E, F$ respectively. Show that

$\frac{A F}{F B} + \frac{A E}{E C} = \frac{A P}{P D}$

HARD

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 58

Two circles with radii $a$ and $b$ touch each other externally. Let $c$ be the radius of the circle that touches these two circles externally as well as a common tangent to the two circles. Prove that $\frac{1}{\sqrt{c}} = \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}}$ .