Gyan Room - Geometry - Concepts & Shortcuts

• Number of Triangles with Integer sides for a given perimeter.

• If the perimeter p is even then, total triangles is [p^2]/48.
• If the perimeter p is odd then, total triangles is [(p+3)^2]/48
• If it asks for number of scalene triangle with a given perimeter P, then subtract 6 and apply the same formula . For even [(p-6)^2]/48 and for odd [(p-3)^2]/48.

Where [ x ] represents neatest integer function. For example [6.7] is 7 not 6 because its nearest integer.

Find the number of triangles with exactly one side odd and perimeter = 203

sum is odd so possibilities are
odd + odd + odd
or
odd + even + even

because one side HAS TO BE odd, its given in the question, so only these 2 possibilities.

But again question says that EXACTLY one side has to be odd .. so we remove the case when all sides are odd.
So total triangles possible - triangles with odd sides
total triangles = (n+3)^2/48 = 884
triangles with sides odd = (101 - 2x) + (101 -2y) + (101-2z) = 203
=> x + y + z = 50
unordered solutions = 234
so 234 triangles have all sides odd..
So total triangles with EXACTLY one side odd = 884 - 234 = 650

• Cube problems - shortcuts

A cube is given with an edge of unit N. It is painted on all faces. It is cut into smaller cubes of edge of unit n. How many edges will have x faces painted?

No of smaller cubes (N/n)^3
No of smaller cubes 5 faces painted 0
4 faces painted 0
3 faces painted 8
2 faces painted 12 * (N-2n)/n
1 face painted 6[N-2n/n]^2
0 face painted [(N-2n)/n]^3

A wooden cuboid of dimensions m * n * p unit is painted in a fixed pattern.
(1) Two opposite faces are painted in red.
(2) Two other opposite faces are painted in green
(3) The remaining two faces are painted in blue

The cuboid is cut into mnp smaller cubes
All faces coloured = 8
two faces coloured = 4{(m-2)+(n-2)+(p-2)}
1 face coloured = 2{(m-2)(n-2)+(n-2)(p-2)+(p-2)(m-2)}
No face coloured = (m-2)(n-2) (p-2)

• How many regular polygons can be formed by joining the vertices of a N - sided regular polygon ?

Number of distinct polygons possible = (Factors of N - 2)
Number of indistinct polygons = Sum of factors of N barring N and N/2

Example : There is a 36 sided regular polygon. How many regular polygons can be formed by joining the vertices of the regular polygon ?
N = 36 = 2^2 * 3^2
Number of Factors = 9
Number of distinct polygons possible = 9 - 2 = 7
Sum of factors = (2^0 + 2^1 + 2^2) (3^0 + 3^1 + 3^2) = 7 * 13 = 91
Number of indistinct polygons = Sum of factors of N barring N and N/2 = 91 - (36 + 18) = 37

• Let say radius of small circle (between the three circles) = r
then 1/r = 1/a + 1/b + 1/c

So if a = 3, b = 6, c = 2
then 1/r = 1/3 + 1/2 + 1/6 = 1
r = 1

• Area of the largest rectangle that can be inscribed in a right triangle = 1/4 * Base * Height

Diameter of the circle inscribed in a right triangle of sides a, b and c (c is hypotenuse) is a + b - c

• In a cyclic quadrilateral ABCD let say, AB = a, BC = b, CD = c and DA = d.

Ratio of the length of Diagonals (AC and BD) = (a * d + b * c) /(a * b + d * c)

Example : In a cyclic quadrilateral ABCD, let AB = 2 cm, BC = 3 cm, CD = 4 cm and DA = 5 cm.
Then ratio of the length of diagonals = (2 * 5 + 3 * 4)/(2 * 3 + 5 * 4) = 22/26 = 11/13

(credits : @Deekonda-Saikrishna)

• Number of Triangles in the below figure = a * b * (b - 1) / 2, where a = number of horizontal lines, b = number of non-horizontal lines. Here a = 4, and b = 4. So Total number of triangles = 4 * 4 * 3/2 = 24

[credits : Apratim Roy]

• Number of Triangles in the figures like the one given below = [n*(n+2)(2n+1)/8], where n is the number of levels. Here, n = 4 so number of triangles = 4 * 6 * 9/8 =  = 27 triangles Here, n = 3 so number of triangles = 3 * 5 * 7 / 8 = [13.125] = 13 triangles

[credits : Ratnesh Soni]

• Ceva's Theorem Given a triangle ABC with a point P inside the triangle, continue lines AP, BP and CP to hit BC, CA and AB at D, E and F respectively. Then AF/FB * BD/DC * CE/EA = 1

Example :

Find x assuming blue lines are concurrent x * 3 * 5 = 2 * 7 * 4
x = 56/15

• Mass point Geometry lecture - @gaurav_sharma

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