Geometry and Trigonometry | PRMO 2019 | Problem 11

Try this beautiful problem from Pre RMO, 2019 based on Geometry and Trigonometry.

Geometry and Trigonometry - PRMO 2019

How many distinct triangles ABC are there, up to similarity, such that the magnitudes of angle A, B and C in degrees are positive integers and satisfy cosAcosB + sinAsinBsinkC=1 for some positive integer k, where kC does not exceed 360 degrees.

is 13
is 25
is 6
cannot be determined from the given information

Key Concepts

Geometry

Trigonometry

Number Theory

Check the Answer

Answer: is 6.

PRMO, 2019

Plane Trigonometry by Loney

Try with Hints

Here cosAcosB+sinAsinBsinkC=1 then cosAcosB+sinAsinB+sinAsinBsinkC-sinAsinB=1 then sinAsinB(sinkC-1)=1-cos(A-B)

Then sinkC-1=0 and cos(A-B)=1 then kC=90 and A=B

Then Number of factors of 90 is 90=(2)($3^{2}$)(5) then number of factors=(2)(3)(2)=12 for 6 factor A,B are integers

Hexagon and Triangle |AMC 8- 2015 -|Problem 21

Try this beautiful problem from Geometry based on hexagon and Triangle.

Area of Triangle | AMC-8, 2015 |Problem 21

In The given figure hexagon ABCDEF is equiangular ,ABJI and FEHG are squares with areas 18 and 32 respectively.$\triangle JBK $ is equilateral and FE=BC. What is the area of $\triangle KBC$?

Key Concepts

Geometry

Triangle

hexagon

Check the Answer

Answer:$12$

AMC-8, 2015 problem 21

Pre College Mathematics

Try with Hints

Clearly FE=BC

Can you now finish the problem ..........

$\triangle KBC$ is a Right Triangle

can you finish the problem........

Clearly ,since FE is a side of square with area 32

Therefore FE=$\sqrt 32$=$4\sqrt2$

Now since FE=BC,We have BC=$4\sqrt2$

Now JB is a side of a square with area 18

so JB=$\sqrt18$=$3\sqrt2$. since $\triangle JBK$ is equilateral BK=$3\sqrt2$

Lastly $\triangle KBC$ is a right triangle ,we see that

$\angle JBA + \angle ABC +\angle CBK +\angle KBJ$ =$360^\circ$

i.e$ 90^\circ + 120^\circ +\angle CBK + 60^\circ=360^\circ$

i.e $\angle CBK=90^\circ $

So $\triangle KBC $ is a right triangle with legs $3\sqrt 2$ and $4\sqrt2$

Now its area is $3\sqrt2 \times 4\sqrt 2 \times \frac {1}{2}$=$\frac{24}{2}$=12

Area of a square | AMC 8- 2015| Problem 25

Try this beautiful problem from AMC-8-2015 (Geometry) based on area of square.

Area of a square - AMC 8, 2015 - Problem 25

One-inch Squares are cut from the corners of this 5 inch square.what is the area in square inches of the largest square that can be fitted into the remaining space?

Key Concepts

Geometry

Area

Square

Check the Answer

Answer:15

AMC-8, 2015 problem 25

Challenges and Thrills of Pre College Mathematics

Try with Hints

Find the Length of HG......

Can you now finish the problem ..........

Draw the big square in the remaining space of the big sqare and find it's area .......

can you finish the problem........

We want to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base 3 and height 1 . so the combined area of the four triangles is $ 4 \times \frac {3}{2} $=6.

The area of the smaller square is 9+6=15.

Radius of a Semicircle | AMC 8, 2016 | Problem 25

Try this beautiful problem from Geometry based on Radius of a semicircle inscribed in an isosceles triangle.

Radius of a Semi circle - AMC-8, 2016 - Problem 25

A semicircle is inscribed in an isoscles triangle with base 16 and height 15 so that the diameter of the semicircle is contained in the base of the triangle as shown .what is the radius of the semicircle?

$\frac{110}{19}$
$\frac{120}{17}$
$\frac{9}{5}$

Key Concepts

Geometry

Area

pythagoras

Check the Answer

Answer:$\frac{120}{17}$

AMC-8, 2016 problem 25

Challenges and Thrills of Pre College Mathematics

Try with Hints

Draw a perpendicular from the point C on base AB

semicircle inscribed in an isosceles triangle

Can you now finish the problem ..........

D be the midpoint of the AB(since $\triangle ABC $ is an isoscles Triangle)

Find AC and area

can you finish the problem........

radius of a semicircle in an isosceles triangle

Area of the $\triangle ABC= \frac{1}{2} \times AB \times CD$

= $ \frac{1}{2} \times 16 \times 15 $

=120 sq.unit

Using the pythagoras th. $ AC^2= AD^2+CD^2$

i.e $AC^2=(8)^2+(15)^2$

i.e $AC=17$

Let$ ED = x$ be the radius of the semicircle

Therefore Area of $\triangle CAD = \frac{1}{2} \times AC \times ED$=$\frac {1}{2} area of \triangle ABC$

i.e $\frac{1}{2} \times AC \times ED $=60

i.e $\frac{1}{2} \times 17 \times x$ =60

i.e $x=\frac {120}{7}$

Area of a Triangle -AMC 8, 2018 - Problem 20

Try this beautiful problem from Geometry based on Area of a Triangle Using similarity

Area of Triangle - AMC-8, 2018 - Problem 20

In $\triangle ABC $ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.

What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?

$\frac{2}{3}$
$\frac{4}{9}$
$\frac{3}{5}$

Key Concepts

Geometry

Area

similarity

Check the Answer

Answer:$\frac{4}{9}$

AMC-8, 2018 problem 20

Pre College Mathematics

Try with Hints

$\triangle ADE$ $\sim$ $\triangle ABC$

Can you now finish the problem ..........

$\triangle BEF$ $\sim$ $\triangle ABC$

can you finish the problem........

Since $\triangle ADE$$\sim$ $\triangle ABC$

$\frac{ \text {area of} \triangle ADE}{ \text {area of} \triangle ABC}$=$\frac{AE^2}{AB^2}$

i.e $\frac{\text{area of} \triangle ADE}{\text{area of} \triangle ABC}$ =$\frac{(1)^2}{(3)^2}$=$\frac{1}{9}$

Again $\triangle BEF$ $\sim$ $\triangle ABC$

Therefore $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$=$\frac{BE^2}{AB^2}$

i.e $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$ =$\frac{(2)^2}{(3)^2}$=$\frac{4}{9}$

Therefore Area of quad. CDEF =$\frac {4}{9}$ of area $\triangle ABC$

i.e The ratio of the area of quad.CDEF to the area of $\triangle ABC$ is $\frac{4}{9}$

Area of cube's cross section |Ratio | AMC 8, 2018 - Problem 24

Try this beautiful problem from Geometry: Ratio of the area of cube's cross section . You may use sequential hints to solve the problem.

Area of cube's cross section - AMC-8, 2018 - Problem 24

In the cube ABCDEFGH with opposite vertices C and E ,J and I are the mid points of segments FB and HD respectively .Let R be the ratio of the area of the cross section EJCI to the area of one of the faces of the cube .what is $R^2$ ?

$\frac{5}{4}$
$\frac{3}{2}$
$\frac{4}{3}$

Key Concepts

Geometry

Area

Pythagorean theorem

Check the Answer

Answer:$\frac{3}{2}$

AMC-8(2018) Problem 24

Pre College Mathematics

Try with Hints

EJCI is a rhombus by symmetry

Can you now finish the problem ..........

Area of rhombus is half product of its diagonals....

can you finish the problem........

Let Side length of a cube be x.

then by the pythagorean theorem$ EC=X \sqrt {3}$

$JI =X \sqrt {2}$

Now the area of the rhombus is half product of its diagonals

therefore the area of the cross section is $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$

This shows that $R= \frac{\sqrt6}{2}$

i.e$ R^2=\frac{3}{2}$

Geometry of Plane figures | Pre-RMO 2019

Try this beautiful problem from Pre-RMO, 2019 based on Geometry of plane figures.

Geometry of Plane figures - Pre-RMO 2019

From a square with sides of length 5, triangular pieces from the four corners are removed to form a regular octagon. Find the area removed to the nearest integer.

Key Concepts

Area of Plane figures

Algebra

Approximation in number theory

Check the Answer

Answer: 4.

Pre-RMO, 2019

Geometry Revisited by Coxeter .

Try with Hints

A figure is taken such that side of triangle is x and sides of octagon are $x\sqrt 2$ and $5-2x$

Here $5-2x$=$x\sqrt 2$ then x=$\frac{5}{2+\sqrt 2}$

Area removed =2x^2=$\frac{2(25)}{2(1+2^\frac{1}{2})}$ which is nearly 4.3 then the nearest integer is 4

Radius of a Semi Circle -AMC 8, 2017 - Problem 22

Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.

AMC-8(2017) - Geometry (Problem 22)

In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?

$\frac{7}{6}$
$\frac{10}{3}$
$\frac{9}{8}$

Key Concepts

Geometry

congruency

similarity

Check the Answer

Answer:$\frac{10}{3}$

AMC-8(2017)

Pre College Mathematics

Try with Hints

Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.

Can you now finish the problem ..........

Now the $\triangle ODB $and $\triangle OCB$ are congruent

can you finish the problem........

Let x be the radius of the semi circle

Now the $\triangle ODB$ and $\triangle OCB$ we have

OD=OC

OB=OB

$\angle ODB$=$\angle OCB$= 90 degree`

so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)

BD=BC=5

And also $\triangle ODA$ and $\triangle BCA$ are similar....

$\frac{8}{12}$=$\frac{x}{5}$

i.e x =$\frac{10}{3}$

Radius of a Circle - SMO 2013 - Problem 25

Try this beautiful problem from Geometry based on the radius and tangent of a circle.

SMO 2013 - Geometry (Problem 25)

As shown in the figure below ,circles $C_1 $and$ C_2$ of radius 360 are tangent to each other , and both tangent to the straight line l.if the circle$ C_3$ is tangent to $C_1$ ,$C_2$ and l ,and circle$ C_4 $is tangent to$ C_1$,$C_3$ and l ,find the radius of$ C_4$

Key Concepts

Geometry

Pythagoras theorm

Distance Formula

Check the Answer

Answer:40

SMO -Math Olympiad-2013

Pre College Mathematics

Try with Hints

Let R be the radius of $C_3$

$C_2E$ =360-R

$C_3E=360$

$C_2C_3$=360+R

Using pythagoras theorm ....

$ (360-R)^2+360^2=(360+R)^2$

i.e R=90

Can you now finish the problem ..........

Let the radius of$ C_4$ be r

then use the distacce formula and tangent property........

can you finish the problem........

Let r be the radius of $C_4$ (small triangle).

LO+OC=360

$\sqrt{(360+p)^2-(360-p)^2}+\sqrt{(90+r)^2-(90-r)^2}=360$

i.e r=40.

Cubes and Rectangles | Math Olympiad Hanoi 2018

Cubes and Rectangles - Math Olympiad Hanoi 2018

Try this beautiful problem from Math Olympiad Hanoi, 2018 based on Cubes and Rectangles.

Find the number of rectangles can be formed by the vertices of a cube.

Key Concepts

Geometry

Permutation

Combination

Check the Answer

Answer: 12.

Math Olympiad Hanoi 2018

Geometry Vol I to IV by Hall and Stevens

Try with Hints

There are 6 squares on 6 faces on the cube.

There are 4 diagonals of the cube that have the same length

and pass through the center of the cube. Every two diagonals intersect at the midpoint and form a rectangle.

Then there are 6+ $(\frac{4!}{2!2!})$ =12 rectangles.