Mathematics Dictionary

It is hard to rember the geometric method
But coordinate geometric method is straightfoward
- We know the slope of three sides
- We can find the equation of three heights
- We can find intesections of height 1 and height 2 at (u1, v1)
- We can find intesections of height 2 and height 3 at (u2, v2)
- We found that u1 = u2 and v1 = v2
- Hence they are concurrent

From the example we see that we just rember the procedures to find solution

Subject | Compare with geometric method

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AN 11 02. Techniques in coordinate geometry

Distance between two points

D = Sqr((x2-x1)^2 + (y2-y1)^2)

Slope between two points

S = (y2-y1)/(x2-x1) and x2 NE x1

Slope relation between two lines

Line 1 : y = s1*x + b1
line 2 : y = s2*x + b2
If line 1 is parallel to line 2 then s1 = s2
if line 1 is perpendicular to line 2 then s1 = -1/s2

Equation of line for giving slope S and point (x1,y1)

Equation is y = s*x + b
Since (x1,y1) is on line, hence y1 = s*x1 + b
Then b = y1 - s*x1
Hence equation of line is y = s*x + y1 - s*x1

Intersection of two lines

Line 1 : y = s1*x + b1
line 2 : y = s2*x + b2
Find point of intesection (u,v)
- s2*u + b2 = s1*u + b1
- Hence u = (b1 - b2)/(s2 - s1)
- Hence v = s1*u + b1

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AN 11 03. In-center of triangle

Let AB be horizontal. Let A(0,0), B(b,0), C(x3,y3)
Slopes
- Slope of bisector of angle A is s1 = tan(A/2)
- Slope of bisector of angle B is s2 = tan(B/2)
- Slope of bisector of angle C is s3 = tan(C/2)
Equations of bisectors
- y = s1*x + y1 - s1*x1
- y = s2*x + y2 - s2*x2
- y = s3*x + y3 - s3*x3
Intersections of bisectors
- Bisector 1 and bisector 2 is at (u1,v1)
- Bisector 2 and bisector 3 is at (u2,v2)
- u1 = u2 and v1 = v2
Hence they are concurrent at I which is called in-center
- We know coordinate of in-center I(u1,v1)
- In-radius r is the distance from I to AB
- In this case r = v1
- Hence we can draw a circle with 3 sides as tangent to circle

Coordinate of in-center

Let side AB be in horiztotal direction
Let BC = a, CA = b and AB = c
Tangent from A to in-circle is AF = (s - a) where IF is distance from I to AB
Coordinate of in-center is I(h,k)
- h = (s - a) where s = (a+b+c)/2 and a,b,c are sides of triangle
- k = (s - a)*tan(A/2)

Diagram

Subject | Diagram 11 03

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AN 11 04. Circum-center of triangle

Let AB be horizontal. Let A(0,0), B(b,0), C(x3,y3)
Slopes
- Slope of bisector of side AB is s1 = 0 and bisector of side AB is infinite
- Slope of bisector of side BC is s2 and bisector of side BC is s22 = -1/s2
- Slope of bisector of side CA is s3 and bisector of side CA is s33 = -1/s3
- Slope of bisector of angle C is s3 = tan(C/2)
Equations of bisectors :
Intersection of bisector 1 and bisector 2 is at (u1,v1)
Intersection of bisector 2 and bisector 3 is at (u2,v2)
u1 = u2 and v1 = v2
Hence they are concurrent at U which is called circum-center
- Circum-radius is UA
- We can draw a circle passing point A, B, C

Coordinate of circum-center

Let side AB be in horiztotal direction
Let BC = a, CA = b and AB = c
Let D be mid-point of BC, E be mid point CA and F be mid point of AB
Coordinate of circum-center is I(h,k)
- h = AF where F is mid point of AB
- k = r = a/(2*sin(A)) by sine law a = 2*r*sin(A)

Diagram

Subject | Diagram 11 03

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AN 11 05. Ex-center of triangle

Let AB be horizontal. Let coordinates be A(0,0), B(b,0), C(x3,y3)
Slopes
- Slope of bisector of internal angle A is s1 = tan(A/2)
- Slope of bisector of external angle B is s2 = tan((pi - B)/2)
- Slope of bisector of external angle C is s2 = tan((pi - C)/2)
Equations of bisectors
- y = s1*x + y1 - s1*x1
- y = s2*x + y2 - s2*x2
- y = s3*x + y3 - s3*x3
Intersections
- Bisector 1 and bisector 2 is at (u1,v1)
- Bisector 2 and bisector 3 is at (u2,v2)
- u1 = u2 and v1 = v2
Hence they are concurrent at I which is called ex-center
- We know coordinate of ex-center I(u1,v1)
- In-radius r is the distance from I to AB
- In this case r = v1 (Assume AB is x-axis)
- Hence we can draw a circle with 3 sides as tangent to ex-circle

Ex-central triangle

Draw 3 ex-centers J, K, and L of triangle ABC
Triangle JKL is called ex-central triangle
Incenter of triangle ABC is at same position of orthocenter of triangle JKL

Coordinate of ex-center

Let side AB be in horiztotal direction
Let BC = a, CA = b and AB = c
Coordinate of circum-center is J(h,k) between sides AB and AC
- h = AF where F is AB tangenting the ex-circle = s and s = (a+b+c)/2
- k = r = AF*tan(A/2) = s*tan(A/2)

Diagram

Subject | Diagram 11 03

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AN 11 06. Centroid and gravity-center of triangle

Let AB be horizontal. Let coordinates be A(x1,y1), B(x2,y2), C(x3,y3)
Mid points of medians
- Median AD : Med point of BC is D(xd,yd) where xd=(x2+x3)/2 and yd=(y2+y3)/2.
- Median BE : Med point of BC is E(xe,ye) where xe=(x3+x1)/2 and ye=(y3+y1)/2.
- Median CF : Med point of BC is F(xf,yf) where xf=(x1+x2)/2 and yf=(y1+y2)/2.
Slopes of medians
- s1 = (yd-y1)/(xd-x1)
- s2 = (ye-y2)/(xe-x2)
- s3 = (yf-y3)/(xf-x3)
Equations of medians
- y = s1*x + y1 - s1*x1
- y = s2*x + y2 - s2*x2
- y = s3*x + y3 - s3*x3
Intersections
- Bisector 1 and bisector 2 is at (u1,v1)
- Bisector 2 and bisector 3 is at (u2,v2)
- u1 = u2 and v1 = v2
Hence they are concurrent at G which is called gravity-center or centroid
- We know coordinate of ex-center G(u1,v1), then we can prove
- AG = 2*AD/3 and GD = AD/3
- BE = 2*BE/3 and GE = BE/3
- CG = 2*CF/3 and GF = CF/3
- In this case r = v1 (Assume AB is x-axis)
- Hence we can draw a circle with 3 sides as tangent to ex-circle

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AN 11 07. Ortho-center of triangle

Let AB be horizontal. Let coordinates be A(x1,y1), B(x2,y2), C(x3,y3)
Slopes of altitudes
- Altitude AD : s1 = -1/S23 and s23 = (y3-y2)/(x3-x2)
- Altitude BE : s2 = -1/S31 and s31 = (y1-y3)/(x1-x3)
- Altitude CF : s3 = -1/S12 and s31 = (y2-y2)/(x2-x1)
Equations of medians
- y = s1*x + y1 - s1*x1
- y = s2*x + y2 - s2*x2
- y = s3*x + y3 - s3*x3
Intersections
- Bisector 1 and bisector 2 is at (u1,v1)
- Bisector 2 and bisector 3 is at (u2,v2)
- u1 = u2 and v1 = v2
Hence they are concurrent at G which is called ortho-center or centroid
- We know coordinate of ex-center O(u1,v1), then we can prove
- A,B,D,E are concyclic if draw circle using AB as diameter
- B,C,E,F are concyclic if draw circle using BC as diameter
- C,A,F,D are concyclic if draw circle using CA as diameter

Pedal triangle

The feet of altitudes make a triangle DEF which is called Pedal triangle

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AN 11 08. Slope and area of triangle

diagram

Subject | Diagram 11 11

Construction

Draw OXY coordinate
Draw point A(0,y) and C(0,v) on y-axis. Let y GT v
Draw point B(x,0) and D(u,0) on x-axis. Let u GT x
Join AB and CD. Let AB and CD meet at E(4,6)

Question

Area of triangle AOB = 54 and Area of traingle OCD = 48. O is the origin
Find area of triangle EFD where F is a point at (4,0)

Solution

1. x*y = 2*54 = 108
2. u*v = 2*48 = 96
3. Sope along AB
- Slope of AE = (y - 6)/(0 - 4)
- Slope of EB = (6 - 0)/(4 - x)
- Hence -(y - 6)/4 = 6/(4 -x)
- Or -(y - 6)*(4 - x) = 24
- or x*y - 6*x - 4*y = 0
4. Eliminate y from (1) and (3)
- x^2 - 18*x + 72 = 0
- (x - 6)*(x - 12) = 0
- Hence x = 6 or x = 12
- Substitute x into (1) we get y = 18 or 9
5. Sope along CD
- Slope of CE = (v - 6)/(0 - 4)
- Slope of ED = (6 - 0)/(4 - u)
- Hence -(v - 6)/4 = 6/(4 - u)
- Or -(v - 6)*(4 - u) = 24
- or u*v - 6*u - 4*v = 0
6. Eliminate v from (2) and (5)
- u^2 - 24*u + 144 = 0
- (u - 12)*(u - 12) = 0
- Hence u = 12 or u = 12
- Substitute u into (2) we get v = 8 or 8
Area of triangle EBD
- Area of EBD = EF*BD/2 = 6*(v - x)/2 = 6*(8 - 6)/2 = 6
Area of triangle EFD
- Area of EBD = EF*FD/2 = 6*(v - 4)/2 = 6*(8 - 4)/2 = 12

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AN 11 09. Distance between orhto-center and circum-center

diagram

Subject | Diagram 11 10

Geometric and trigonometric method

Construction
- Draw triangle ABC. Let O be the ortho-center and Q be the circum-center
- Let COH is the height
- Join CD and draw DE perpendicular to AC
Triangle AQD : Angle QAD = 90 - B
- Angle CDE = 90 - CDE
- Use circum-circle, we have angle CDE = angle B
- Hence Angle QAD = 90 - B
Triangle HAB : Angle OCA = 90 - A
Hence angle OCD = angle OCA - angle DCE = B - A
DC = R
OC = 2*R*cos(C)
Triangle DCO and we use cosine law
- OD^2 = DC^2 + OC^2 - 2*DC*OC*cos(DCO)
- OD^2 = R^2 + (2*R*cos(C))^2 - 4*(R^2)*cos(C)*cos(OAQ)
- OD^2 = R^2 + 4*(R^2)*cos(C)*(cos(A) - cos(B-A))
- OD^2 = R^2 + 4*(R^2)*cos(C)*(-cos(A+B)-cos(B-A))
- OD^2 = R^2 + 8*(R^2)*cos(A)*cos(B)*cos(C)

Analytic geometric method

Find ortho-center O using intersection of two heights
Find circum-center D using intersection of two side bisectors
Use distance formula to find OD

Verify

Use data in diagram to verify OD = (R^2)*(1-8*cos(A)*cos(B)*cos(C)
Where a = 2*R*sin(A)

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Q10. Triangle ABC has four triangle by given points on side

Construction

Draw a triangle ABC and sides are a,b,c
Let E be a point on AC and CE = 2*b/3 and AE = b/3. AC = b
Let T be a point on BA and BA = c
Let F be a mid point of BC and BC = a
Join ET, TF and FE so that make four triangles
Let area of triangle BTF = x
Let area of triangle EFC = y
Let area of triangle ATE = z

Question

If x^2 = y*z, find the ratio BT : TA
Find the area of triangle ETF

Solution

Let points be B(0,0), C(4,0) and A(x1=3,y1=3)
y = Area EFC
- Area = FC*h/2
- FC = a/2 and h = 2*y1/3 = 2. a = BC = 4
- Area = (a/2)*(2)/2 = (4/2)*(2)/2 = 2
x = Area BFT and T(u,v)
- Area = v*(a/2)/2 = v*(4/2)/2 = v
z = area AET
- | x1 y1 1 |
- | xE yE 1 | = x1*yE + y1*xT + yT*xE - xT*yE - xE*y1 - x1*yT
- | xT yT 1 |
- x1 = 3, y1 = 3
- xT = u, yT = v
- XF = 2, yF = 0
- Hence area = (4 - u - v/3)/2
x^2 = y*z
- v^2 = (2)*(4 - u - v/3)/2
- v^2 = 4 - u - v/3
- 3*(v^2) + v + (3*u - 12) = 0
BT + TA = c
- Sqr(u^2 + v^2) = Sqr((u-3)^2 + (v-3)^2)
- u^2 + v^2 = u^2 - 6*u + 9 + v^2 - 6*v + 9
- 6*(u + v) = 18 or u + v = 3
Solve 3*(v^2) + v + (3*u - 12) = 0 and (u + v) = 3
- 3*v^2 + v + (3*(3 - v) - 12) = 0
- 3*v^2 + v + 9 - 3*v - 12 = 0
- 3*v^2 - 2*v - 3 = 0
- v = (2 + Sqr(4 + 4*3*3)))/6 or v = (2 - Sqr(4 + 4*3*3))/6
- v = (2 + 6.32455)/6 or v = (2 - 6.32455)/6
- v = 1.387425
- u = 3 - v = 1.6125741
BT
- BT = Sqr(u^2 + v^2)
- BT = Sqr ((1.6125741)^2 + (1.38745)^2)
- BT =
TA
- TA = Sqr((u-3)^2 + (v-3)^2)
- TA = Sqr((3-1.6125741)^2 + (3-1.38745)^2)
- TA =

Verify

Find y*z
- y = 2
- z = (4 - u - v/3)/2
- Hence y*z = (4 - u - v/3)
- = 4 - 1.6125741 - 1.387425/3
- = 2.3874259 - 0.462475 = 1.924951
Find x^2
- x^2 = v^2 = (1.387426)^2 = 1.924951

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Q11. Triangle ABC :

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Q12. Triangle : Ortho-center, centroid and circum center are colinear

Diagram

Subject | Diagram GE 27 15

Geometric proof

See GE 16 12

Analytic geometric proof

Give coordinates of point A, B, C of triangle ABC
Find coordinate of ortho-center O
- Find intersection of two heights
Find coordinate of circum center V
- Find intersection of two bisectors of sides
Find coordinate of centroid G
- Find intersection of two medians
Find slope of OG and OV
If slope OG = slope OV, then the proof is completed

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Q00. Outlines
Techniques in coordiante geometry

Find slope from two points
Find equation of lines by slope and point method
Find intersection of two lines
Find distance between two points

Prove five centers of triangle by coordinate geometry

In-center : angle bisectors are concurrent
Circum-center : side bisectors are concurrent
Ex-center : angle bisectors are concurrent (one interanl and two external)
Centroid : Medians are concurrent
Ortho-center : Altitudes are concurrent

Coordinates of five centers of triangle

In-center : h = (s-a) and k = (s-a)*tan(A/2)
Circum-center : h = c/2 and k = a/(2*sin(A))
Ex-center : h = s and k = s*tan(A/2)
Centroid :
Ortho-center :

Questions

AL 11 08 : Slope and area of triangle

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