Formulae

Question 1

Area of a triangle with sides a, b and included angle θ

Answer 1

1/2 · a · b · sin θ

Question 2

a, b, c are the sides of the triangle with circumradius R.

Answer 2

abc/4R

Proof

Question 3

Area of a kite

2.2 cm wide and 8 cm long

Answer 3

• d₁d2
  8. 8 cm²</sub>

Question 4

Area of cyclic quadrilateral

Answer 4

1. s = (a + b + c)/2
2. sqrt( (s-a) * (s-b) * (s-c))

where s is the semiperimeter and a, b and c are the side lengths.

Question 5

Area of triangle with sides a, b and c

Answer 5

1. Let s= (a + b + c)/2
2. Area = sqrt (s * (s-a) * (s-b) * (s-c))

Heron’s formula:

Question 6

Area of regular polygon with perimeter p and apothem a.

Answer 6

• pa/2*
• p·cot 180/n* (where n is the number of sides)

1. Inscribe the polygon a circle
2. Draw a line from two adjacent vertices to the circumcenter.
3. This creates a triangle that is 1/n of the total area.

Question 7

Apothem: length of

Answer 7

The apothem of a regular polygon is the line segment drawn from the center of the polygon perpendicular to one of its edges. It is also the radius of the inscribed circle of the polygon.

Given the number of sides n and side length s the length of the apothem is s /( 2·tan (π/n) ).

Question 8

Area of a regular hexagon with side s

Answer 8

A = (3/2)* sqrt(3) * s²

Question 9

area of a hexagon with perimeter 24 cm

Answer 9

24*sqrt(3) square cm

~ 41.569 cm sq

Question 10

Volume/Surface Area of a Cone

Answer 10

• V =πr²h/3*
• SA=πr²+πrl*

Where V is the volume, SA is the surface area, r is the radius of the circular base, h is the height, and l is the slant height.

Question 11

Volume/Surface Area of a Sphere*

Answer 11

• V = (4/3 ) · πr³*
• SA = 4πr²*

Where V is the volume, SA is the surface area, and r is the radius of the sphere (which is radius of the central cross section/the base of the semisphere).

Question 12

Volume/Surface Area of a Pyramid

Answer 12

• V = (1/2)bh*
• SA = 2sl + b*

Where V is the volume, SA is the surface area, b is the area of the base, h is the height, l is the slant height, and s is the length of a side of the base. Note that a pyramid can have a base of any polygon, but if none is specified, assume a square base. A pyramid with a triangular base is known as a tetrahedron.

Question 13

Volume/Surface Area of a Prism*

Answer 13

• V =lwh*
• SA=2(lw+lh+wh)*

Where V is the volume, SA is the surface area, l is the length, w is the width, and h is the height.

Question 14

Common Pythagorean triples

Answer 14

3 4 5

5 12 13

7 24 25

8 15 17

Important:or these side lengths multiplied by some factor, it is a right triangle)

Question 15

distance between the line ax+by+c=0 and point (x1,y1) is

Answer 15

• |ax1 + by1+c| /sqrt(a²+b²)*

Question 16

Distance between two points (x₁,y₁) and (x₂, y₂)

Answer 16

In the -dimensional case, the distance between (a₁, a₂ ..an) and (b₁,b₂, ..b_n) is:

sqrt( Σ (a_i-b_i)²)

Question 17

A convex quadrilateral is cyclic if and only if the four perpendicular bisectors to the sides are

Answer 17

Concurrent.

This common point is the circumcenter.

Question 18

Angles of a cyclic quadrilateral ABCD with diagonals

Answer 18

∠A + ∠C = 180°

∠ABD = ∠ACD

∠BCA = ∠BDA

∠BAC = ∠BDC

∠CAD = ∠CBD

Question 19

Ptolemy’s Theorem*

Answer 19

ab + cd = ef

Where ABCD is a cyclic quadrilateral with side lengths a, b, c, d and diagonals e, f (with a opposite b and c opposite d).

Question 20

In ΔABC we have AB=7, AC=8, BC=9. Point D is on the circumscribed circle of the Δ so that AD bisects angle BAC.

What is the value of: AD/CD?

Answer 20

Set BD’s length as x.

CD’s length must also be x since ∠BAD and ∠ DAC intercept arcs of equal length(because ∠ BAD= ∠ DAC).

Using Ptolemy’s Theorem, 7x+8x=9.

The ratio is 5/3

◌

Question 21

Brahmagupta’s Formula

Answer 21

K =sqrt ( (s-a)(s -b)(s-c)(s-d) )

Where K is the area and s is the semiperimeter of the quadrilateral with sides a, b, c, d. For this formula to work, the quadrilateral must be cyclic.

Question 22

A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by AB, has length 31.

Find the sum of the lengths of the three diagonals that can be drawn from A.

1991 AIME Problems/Problem 14

Answer 22

384

Let x=AC=BF, y=AD=BE, and z=AE=BD.

Ptolemy’s Theorem on ABCD gives 81y+31·81=xz and Ptolemy on ACDF gives x·z + 81² = y².

Subtracting these equations give

y²-81y-112·81 = 0 ,

and from this y=144.

Ptolemy on ADEF gives 81y+ 81²= z² , and from this z=135.

Finally, plugging back into the first equation gives x=105,

so

x+y+z = 105 + 144 + 135 = 384.

Question 23

A quadrilateral is inscribed in a circle of radius 200*sqrt(2). Three of the sides of this quadrilateral have length 200. What is the length of the fourth side?

Answer 23

1. Use Ptolemy’s
2. Divide all side lengths by 200 to make computation easiers
3. Construct quadrilateral ABCD on the circle with AD being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center O to A, B, C and D. Let the intersection of BD and OC be point E. Notice that BD and OC are perpendicular because BCDO is a kite.
4. set lengths BE=ED equal to x
5. Solve for x using Pythagoras
6. x=sqrt(14)/2
7. By Ptolemy’s Theorem,

• AB*CD+BC*AD = AC*BD=BD*BD=(2*BE)²*
  8. AD=5/2
  9. Multiply by 200 ==> 500

Question 24

What are the 3 cases of the Power of a Point theorem?

Answer 24

The two lines are secants of the circle and intersect inside the circle (figure on the left). In this case, we have AE*CE=BE*DE

One of the lines is tangent to the circle while the other is a secant (middle figure). In this case, we have AB²=BC*BD

Both lines are secants of the circle and intersect outside of it (figure on the right). In this case, we have CB*CA=CD*CE

Question 25

Power of Point Theorem: Compact formulation

Answer 25

Consider a circle O and a point P which is not on the circle.

Draw a line through P that intersects the circle in two places.

The product of the length from P to the first point of intersection and the length from P to the second point of intersection is constant for any choice of a line through P that intersects the circle.

This constant is called the power of point P.

For example, in the figure below

Question 26

Chords AB and CD of a given circle are perpendicular to each other and intersect at a right angle. Given that BE=16, DE=4 and AD=5 find CE.

Answer 26

1. ADE is a right triangle with hypotenuse 5 and leg 4.
2. Therefore AE=3.
3. Applying t Power of a Point Theorem gives AE*BE=CE*DE or 3*16=x*4

=> x = 12

Question 27

Ceva’s theorem

Answer 27

Criterion for the concurrence of cevians in a triangle.

1. Let ABC be a triangle
2. Llet D, E, F be points on lines BC, CA, AB respectively. Lines AD, BE, CFare concurrent if and only if:

BD/DC · CE/EA · AF/FB = 1

where lengths are directed. This also works for the reciprocal of each of the ratios, as the reciprocal of 1 is 1.

Question 28

Ceva’s theorem: idea behind proof

Answer 28

1. Assume that the cevians are concurrent
2. Show the Ceva property follows using ratios of areas of triangles.
3. Concoct up a point F’ and show that F’ is the same as F.

Suppose AD, BE, CF meet at a point X.

Triangles ABD, ADC have the same altitude to line BC, but bases BD and DC.

It follows that BD/DC = [ABD]/[ADC] . The same is true for triangles XBD, XDC

Suppose D, E and F satisfy Ceva’s criterion, and suppose AD, BE intersect at X. Suppose the line CX intersects line AB at F’. We have proven that F’ must satisfy Ceva’s criterion.

=> AF’/F’B = AF/FB’ so

=> F’=F ,

=> CF concurs with AD and BE.

Question 29

Menelaus’ Theorem

Answer 29

IFF P, Q, R on the respective sides BC, CA, AB (or their extensions) of a triangle ABC are collinear:

BP * CQ * AR = -PC * QA * RB

where all segments in the formula are directed segments.

Question 30

Menelaus’ Theorem: idea behind proof

Answer 30

Draw a line parallel to QP through A to intersect BC at K:

Question 31

An angle formed by three points on the circumference of the circle is equal to _ the measure of the subtended arc.

Answer 31

1/2

Question 32

sin 2θ

Answer 32

2 sin θ cos θ

Question 33

sin (α ± β)

Answer 33

sin α cos β ± sin β cos α

Question 34

Sum of interior angles in a polygon

Answer 34

(n-2)_{<span><strong>·</strong></span>}180°

Question 35

Area of a polygon of inradius r and perimeter p

Answer 35

s*r

where s is the semi-perimeter

1. Add incircle, drop altitudes from the incenter I.
2. Draw the lines AI, BC and CI to make three sub-triangles AIB, AIC, BIC
3. AB, AC and BC are the bases of these triangles with the same height r.
4. => [ABC] = ar/2 + br/2 + cr/2 = sr where s is the semiperimeter

Question 37

The sum of the exterior angles in a polygon:

Answer 37

360^o