The angles in each of two triangles are in G.P. The least angle of one of them is three times the least angle in the other, and the sum of the greatest sngles is 4?/3. Find the circular mesures of the angles.
Question no. 23
Let the angles of the first triangle be a, b, c, in order of size, and the angles of the second triangle be A, B, C, also ordered by size. Note that a, b, c, A, B, C > 0 as these are angles of triangles.
Then we are given:
ac = b²
AC = B²
A = 3a
c + C = 4π/3
a + b + c = π [ angles of a triangle ]
A + B + C = π [ angles of a triangle ]
0 < A ≤ π/3 [ not actually given, but justified in following paragraph ]
As A+B+C=π with A,B,C > 0 and A being the smallest, it follows that 0 < A ≤ π/3.
From 6 variables down to 4 variables - Eliminating A and C
Using A = 3a and C = 4π/3 - c to eliminate A and C from the above gives:
ac = b²
a + b + c = π
(3a)(4π/3 - c) = B² => 4πa - 3ac = B²
3a + B + (4π/3 - c) = π => B = c - 3a - π/3
0 < a ≤ π/9
From 4 variables down to 2 variables - Eliminating B and b
Using the 2rd and 4th of these to eliminate B and b gives:
ac = ( π - a - c )² => ( a - π + c/2 )² = c ( π - 3c/4 )
4πa - 3ac = ( c - 3a - π/3 )² => ( 3a - π/3 - c/2 )² = c ( π - 3c/4 )
0 < a ≤ π/9
From the first two of these we have a - π + c/2 = ±( 3a - π/3 - c/2 ).
Case I : a - π + c/2 = - ( 3a - π/3 - c/2 )
=> 4a = 4π/3 => a = π/3. But this contradicts a ≤ π/9, so this does not lead to a solution.
Case II : a - π + c/2 = 3a - π/3 - c/2
=> c = 2a + 2π/3
Putting this into ac = ( π - a - c )² gives
a ( 2a + 2π/3 ) = ( π - a - 2a - 2π/3 )²
=> 7a² - 8(π/3)a + (π/3)² = 0
=> ( 7a - π/3 ) ( a - π/3 ) = 0
=> either a = π/21 or a = π/3.
As a = π/3 contradicts a ≤ π/9, the only possibility is a = π/21.
Then:
c = 2a + 2π/3 = 2π/21 + 14π/21 = 16π/21
b² = ac = 16π²/21² => b = 4π/21
A = 3a = π/7
C = 4π/3 - c = 28π/21 - 16π/21 = 12π/21 = 4π/7
B² = AC = 4π²/7² => B = 2π/7.
Checking the original conditions are satisfied, the first 4 are since we used them to determine b, A, C and B. So we just need to check that a+b+c=π and A+B+C=π. This is straightforward and checks out okay. So the solution is:
a = π / 21, b = 4π / 21, c = 16π / 21
A = π / 7, B = 2π / 7, C = 4π / 7
Then we are given:
ac = b²
AC = B²
A = 3a
c + C = 4π/3
a + b + c = π [ angles of a triangle ]
A + B + C = π [ angles of a triangle ]
0 < A ≤ π/3 [ not actually given, but justified in following paragraph ]
As A+B+C=π with A,B,C > 0 and A being the smallest, it follows that 0 < A ≤ π/3.
From 6 variables down to 4 variables - Eliminating A and C
Using A = 3a and C = 4π/3 - c to eliminate A and C from the above gives:
ac = b²
a + b + c = π
(3a)(4π/3 - c) = B² => 4πa - 3ac = B²
3a + B + (4π/3 - c) = π => B = c - 3a - π/3
0 < a ≤ π/9
From 4 variables down to 2 variables - Eliminating B and b
Using the 2rd and 4th of these to eliminate B and b gives:
ac = ( π - a - c )² => ( a - π + c/2 )² = c ( π - 3c/4 )
4πa - 3ac = ( c - 3a - π/3 )² => ( 3a - π/3 - c/2 )² = c ( π - 3c/4 )
0 < a ≤ π/9
From the first two of these we have a - π + c/2 = ±( 3a - π/3 - c/2 ).
Case I : a - π + c/2 = - ( 3a - π/3 - c/2 )
=> 4a = 4π/3 => a = π/3. But this contradicts a ≤ π/9, so this does not lead to a solution.
Case II : a - π + c/2 = 3a - π/3 - c/2
=> c = 2a + 2π/3
Putting this into ac = ( π - a - c )² gives
a ( 2a + 2π/3 ) = ( π - a - 2a - 2π/3 )²
=> 7a² - 8(π/3)a + (π/3)² = 0
=> ( 7a - π/3 ) ( a - π/3 ) = 0
=> either a = π/21 or a = π/3.
As a = π/3 contradicts a ≤ π/9, the only possibility is a = π/21.
Then:
c = 2a + 2π/3 = 2π/21 + 14π/21 = 16π/21
b² = ac = 16π²/21² => b = 4π/21
A = 3a = π/7
C = 4π/3 - c = 28π/21 - 16π/21 = 12π/21 = 4π/7
B² = AC = 4π²/7² => B = 2π/7.
Checking the original conditions are satisfied, the first 4 are since we used them to determine b, A, C and B. So we just need to check that a+b+c=π and A+B+C=π. This is straightforward and checks out okay. So the solution is:
a = π / 21, b = 4π / 21, c = 16π / 21
A = π / 7, B = 2π / 7, C = 4π / 7