Piyush Theorem
A New Theorem introduced by Piyush Goel with Four Proofs
Mathematics for Piyush is a Passion from his childhood he was so passionate about Mathematics used to play with Numbers draw figures and try to get sides distance one day I draw a AP SERIES Right Angle Triangle( thinking that the distance between the point of intersection of median & altitude at the base must be sum of rest sides that was in My Mind).
And at last Piyush Succeed. This new Theorem proved with Four Proof(Trigonometry/Co-ordinates Geometry/Acute Theorem/Obtuse Theorem).
Here are the Proofs:
This Theorem applies in Two Conditions:
- The Triangle must be Right-Angled.
- Its Sides are in A.P. Series.
- Proof with Trigonometry
- Proof with Obtuse Triangle Theorem
- Proof with Acute Triangle Theorem
- Proof with Co-ordinates Geometry
Four Proof ( TRIGONOMETRY/CO-ORDINATES/OBTUSE TRIANGLE/ACUTE TRIANGLE)
(By PIYUSH GOEL)
Theorem: In a Right-Angled Triangle with sides in A.P
Theorem: In a Right-Angled Triangle with sides in A.P. Series, the distance between the point of intersection of median & altitude at the base is 1/10 Th the sum of other two sides.
This Theorem applies in Two Conditions:
- The Triangle must be Right-Angled.
- Its Sides are in A.P. Series.
- Proof with Trigonometry
tan α =AD/DC
AD= DC tan α —————–1
tan α = AD/DE
AD= DE tan 2 α —————-2
DC tan α = DE tan 2α
(DE+EC) tan α = DE tan 2α
DE tan α + EC tan α = DE tan 2α
DE tan α + EC tan α = 2 DE tan α / (1- tan^2 α)
DE tan α – DE tan^3 α + EC tan α –EC tan^3 α = 2 DE tan α
EC tan α –EC tan^3 α- DE tan^3 α = 2 DE tan α – DE tan α
tan α (EC – EC tan^2 α – DE t an^2 α )= DE tan α
DE tan^2 α – DE = EC tan^2 α – EC
-DE ( tan^2 α + 1) = -EC (1 – tan^2 α )
DE (sin^2 α /cos^2 α + 1) = EC (1- sin^2 α /cos^2 α )
DE (sin^2 α + cos^2 α /cos^2 α ) = EC (cos^2 α – sin^2 α /cos^2 α )
DE (sin^2 α + cos^2 α ) = EC(cos^2 α -sin^2 α )
DE (sin^2 α + cos^2 α ) = EC (cos^2 α -sin^2 α) ……..where (sin^2 α + cos^2 α =1) & (cos^2 α -sin^2 α = cos 2 α )
DE= EC cos 2 α
cos α =a/a+d & sin α = (a-d)/ (a +d)
cos^2 α = a 2/ (a +b) 2
sin^2 α = (a-d) 2/ (a+ d) 2
DE= EC (cos^2 α – sin^2 α )
= EC ( a 2 / (a +b) 2 – (a-d) 2/ (a +d) 2
= EC ( a 2 – (a-d) 2/ (a +d) 2
= EC (a –a +d) (a+ a-d)/ (a+ d) 2
= EC (d) (2 a -d)/ (a+ d) 2
= (a +d)/2(d) (2 a -d)/ (a +d) 2 ————- where EC= (a +d)/2
= (d) (2 a -d)/2(a +d)
= (d) (8 d -d)/2(4 d+d) ——————where a= 4 d (as per the Theorem)
= 7 d 2 /2 (5 d)
= 7 d /10
= (3 d+4 d)/10= (A B+AC)/10
- Proof with Obtuse Triangle Theorem
AC 2=EC 2 +A E 2 +2 C E. DE where EC = ( a +d) /2,A E=( a +d)/2
a 2 = (a +d/2)2 + (a+ d/2)2 + 2(a +d)/2 DE
= (a +d/2) (a+d+2 DE)
= (a +d/2) (a+d+2 DE) where a= 4 d
16 d 2 = (5 d/2) (5 d+2 DE)
32 d/5 = 5 d + 2 DE
32 d/5 – 5 d = 2 DE
32 d -25 d/5 = 2 DE
DE =7 d /10
= (3 d+4 d)/10 = (A B+AC)/10
- Proof with Acute Triangle Theorem
A B 2= AC 2+BC 2 – 2 BC.DC
(a-d) 2= a 2 + (a+ d) 2 -2(a+ d) (DE+EC) where A B= (a-d), AC=a, BC =( a +d) & EC= (a +d)/2
(a-d) 2 – (a +d)2 = a 2 -2(a +d)(DE+EC)
(a- d –a-d) (a -d +a +d) = a 2 -2(a+ d) (2 DE+a+d)/2
2(-2 d) (2 a) = 2 a 2 -2(a +d) (2 DE+a+d)
-8 ad – 2 a 2 = -2(a +d) (2 DE+a+d)
-2 a (4 d +a) = -2(a +d) (2 DE+a+d)
a (4 d + a) = (a +d)(2 DE+a+d)
4 d (4 d + 4 d) = (4 d+d) (2 DE+4 d+d)
4 d (8 d) = (5 d) (2 DE+5 d)
32 d 2/5 d = (2 DE+5 d)
32 d/5 = (2 DE+5 d)
32 d/5 – 5 d = 2 DE
(32 d – 25 d)/5 = 2 DE
DE = 7 d/10
= (3 d+4 d)/10 = (AB+AC)/10
- Proof with Co-ordinates Geometry
In Triangle A B C,point A,B&C,s co-ordinates are respectively (0,0) ,(a,0)&(0,b).
Point D is middle point ,co-ordinates of Point Dis (a/2,b/2)
Equation of BE is ……………… ( Two Points equation)
Y –Y1 =(X – X1 )(Y2 –Y1)/(X2 – X1 )
Y – 0 =b-0/0-a(X – a)
Y = -b/a(X) + b——————- (1)
M 1 = -b/a
For perpendicular
M1 M2= -1
M2= -1/M1So M2= a/b
Equation of AC
Y – 0 = a/b(X-0)
Y=a/b(X) —————— (2)
Put Y value in equation (1)
a/b(X) + b/a(X) =b
X (a 2+b 2/a b) = b
X = a b 2/ (a 2 + b 2)
To get Value of Y, put X value in equation (2)
Y = a/b (a b 2/ (a 2+b 2)
Y = a 2 b/ (a 2+b 2)
Here we got co-ordinates of Point C – a b 2/ (a 2 + b 2), a 2 b/ (a 2+b 2) and co-ordinates of point D is (a/2, b/2) because d is midpoint.
As per the A.P Series (z-d,z,z+ d)
Hers a= z-d, b= z, c = z+ d
(z +d) 2= (z-d) 2+ z 2
(z +d) 2- (z-d) 2 = z 2
(z +d +z-d)(z +d – z + d) = z 2
(2 z)(2 d) = z 2
4 z d = z 2
4 d = z
Put value of a & b
a b 2/ (a 2 + b 2), a 2 b/ (a 2+ b 2) & (a/2, b/2)
a b 2/ ( a 2 + b 2) = 48 d/25
a 2 b/ (a 2+b 2) = 36 d /25
a/ 2=3 d/2
b/ 2 =4 d/2
CD 2= (48 d/25 -3 d/2)2-(3 6 d/25-4 d/2)2
= (96 d-75 d/50)2 + (72 d-100 d/50)2
= (21 d/50)2 + (-28 d/50)2
= (441 d 2/2500) + (784 d 2/2500)
= (1225 d 2/2500)
CD= 35 d/50 = 7 d/10
= 7 d/10 = (3 d+4 d)/10 = (A B+A E)/10
Copyrighted to PiyushGoel(1987)
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