Given the angle Q B P and a point L outside the angle Q B P , draw a straight line through L meeting B Q in A , B P in C such that the triangle A B C has a given perimeter.

First we analyse the problem. Suppose we have located points A and C such that triangle A B C has the given perimeter, say 2 s . If the ex-circle opposite to B touches the rays B C → and B A → at X and Y . We know that B X = B Y = s . This gives us a method of construction as follows: First locate points X and Y on B P → and B Q → respectively such that B X = B Y = s . Draw the perpendiculars to B P and B Q at X and Y respectively to meet at O . With O as centre and O X (or O Y ) as radius, draw a circle. Draw a tangent from L to the circle to meet the rays B P → and B Q → in C and A respectively. There are two tangents from L to the circle; choose the one nearer to B . Then, A B C is the required triangle.

Triangle A B C has incentre I and the incircle touches B C , C A at D , E respectively. If B I meets D E in G , show that A G is perpendicular to B G .

Join I D and A I . We first show that triangles B D G and B I A are similar. We have ∠ D B G = B 2 = ∠ I B A . . . 1 Further ∆ C D E is isosceles, because C D = C E . So ∠ C D E = ∠ C E D = 90 ° - C 2 Therefore ∠ B D G = 180 ° - ∠ C D E = 90 ° + C 2 . Also ∠ B I A = 180 ° - ∠ A B I - ∠ B A I = 180 ° - B 2 - A 2 = 90 ° + C 2 Thus, ∠ B D G = ∠ B I A . . . 2 From (1) and (2), it follows that, ∆ B D G ~ ∆ B I A A A test of similarity . Therefore B D B I = B G B A This fact along with the relation ∠ D B I = ∠ G B A = B / 2 implies that triangles D B I and G B A are similar. Consequently, ∠ B D I = ∠ B G A . But ∠ B D I = 90 ° . So ∠ B G A = 90 ° . That is, B G is perpendicular to A G . Aliter: Join I E . We show that I G E A is a cyclic quadrilateral. As in the previous method, ∠ C E D = ∠ C D E = 90 ° - C 2 . Therefore ∠ A E G = 180 ° - ∠ D E C = 180 ° - 90 ° - C 2 = 90 ° + C 2 . Also ∠ A I G = ∠ A B I + ∠ B A I = B 2 + A 2 . So, ∠ A E G + ∠ A I G = 90 ° + C 2 + B 2 + A 2 = 180 ° This implies that I G E A is a cyclic quadrilateral. So ∠ A G I = ∠ A E I = 90 ° . That is A G is perpendicular to G I , as required.

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 ° .

Suppose A 1 A 2 A 3 … A n is an n -sided regular polygon such that 1 A 1 A 2 = 1 A 1 A 3 + 1 A 1 A 4 . Determine n , the number of sides of the polygon.

From the given relation, we have A 1 A 2 · A 1 A 3 + A 1 A 2 · A 1 A 4 = A 1 A 3 · A 1 A 4 . . . 1 Also in the cyclic quadrilateral A 1 A 3 A 4 A 5 , by Ptolemy's theorem, we get A 4 A 5 · A 1 A 3 + A 3 A 4 · A 1 A 5 = A 3 A 5 · A 1 A 4 . . . 2 Since A 1 A 2 A 3 … A n is a regular n -sided polygon. So, A 1 A 2 = A 4 A 5 , A 1 A 2 = A 3 A 4 , A 1 A 3 = A 3 A 5 . Thus, equation 2 reduces to A 1 A 2 · A 1 A 3 + A 1 A 2 · A 1 A 5 = A 1 A 3 · A 1 A 4 . . . 3 . Comparing equations 1 & 3 , we get A 1 A 4 = A 1 A 5 Since the two diagonals A 1 A 4 and A 1 A 5 are equal. It follows that there must be the same number of vertices between A 1 and A 4 as between A 1 and A 5 . Since, there are two vertices A 2 & A 3 between A 1 and A 4 . So, there must be two vertices between A 1 and A 5 . Thus, the polygon must be 7 -sided, that is n = 7 . Aliter: If O is the centre of the circle in which A 1 A 2 … A n is inscribed and θ is the angle which each side of the polygon subtends at O , then using the relation 1 A 1 A 2 = 1 A 1 A 3 + 1 A 1 A 4 , obtain an equation in θ . Solve the equation to get θ = 2 π 7 . This means n = 7 .

HARD

Olympiad

IMPORTANT

Earn 100

Construct a triangle $A B C$ given $h_{a}, h_{b}$ , the altitudes from $A$ and $B$ respectively and $m_{a}$ , the median through $A$ .

Important Questions on Geometry

MEDIUM

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 60

Given the angle

Q B P

and a point

L

outside the angle

Q B P

, draw a straight line through

L

meeting

B Q

A

B P

C

such that the triangle

A B C

has a given perimeter.

HARD

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 61

Triangle

A B C

has incentre

I

and the incircle touches

B C, C A

D, E

respectively. If

B I

meets

D E

G

, show that

A G

is perpendicular to

B G

MEDIUM

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 62

Let

A

be one of the two points of intersection of two circles with centres

X

and

Y

. The tangents at

A

to these two circles meet the circles again at

B, C

. Let the point

P

be located so that

P X A Y

is a parallelogram. Show that

P

is the circumcentre of triangle

A B C

MEDIUM

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 63

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 \cdot A B = I B^{2}

and

C Y \cdot A C = I C^{2}

. Given that

X, I, Y

are collinear, find the possible values of the measure of angle

A

HARD

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 64

A triangle

A B C

has incentre

I

. Its incircle touches the side

B C

T

. The line through

T

parallel to

I A

meets the incircle at

S

and the tangent to the incircle at

S

meets sides

A B, A C

in points

C^{'}, B^{'}

respectively. Prove that triangle

A B^{'} C^{'}

is similar to triangle

A B C

MEDIUM

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 65

Suppose $A_{1} A_{2} A_{3} \dots A_{n}$ is an $n$ -sided regular polygon such that $\frac{1}{A_{1} A_{2}} = \frac{1}{A_{1} A_{3}} + \frac{1}{A_{1} A_{4}}$ . Determine $n$ , the number of sides of the polygon.

HARD

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 66

Suppose $A B C D$ is a quadrilateral such that a semicircle with its centre at the midpoint of $A B$ and bounding diameter lying on $A B$ touches the other three sides $B C, C D$ and $D A$ . Show that $A B^{2} = 4 B C \cdot A D$ .

HARD

Olympiad

IMPORTANT

Problem Primer for the Olympiad>Chapter 3 - Geometry>Geometry>Q 67

Let

A B C

be an acute-angled triangle. For any point

P

lying within this triangle, let

D, E, F

denote the feet of the perpendiculars from

P

onto the sides of

B C, C A, A B

respectively. Determine the set of all possible positions of the point

P

for which the triangle

D E F

is isosceles. For which positions of

P

will the triangle

D E F

become equilateral.

Construct a triangle ABC given ha,hb, the altitudes from A and B respectively and ma, the median through A.

Important Questions on Geometry

Learn and Prepare for any exam you want !

Construct a triangle $A B C$ given $h_{a}, h_{b}$ , the altitudes from $A$ and $B$ respectively and $m_{a}$ , the median through $A$ .