In any triangle ABC,
A+B+C = 180 degrees....(a);
0 < sinC ≤ 1...(same for A and B)....(b); and
None of the angles = 0 or 180 deg. ....(c).
For any 2 angles X and Y,
cosXcosY + sinXsinY = cos(X-Y).....(d);
sin²X + cos²X = 1.....(e); and
all sines and cosines are ≤ 1.....(f).
Necessity:
Let's start with condition 1.
sinAsinBsinC + cosAcosB = 1.
sinAsinBsinC ≤ sinAsinB from (b) above.
So, 1 ≤ sinAsinB + cosAcosB.
From (d) above, 1 ≤ cos(A-B).
But 1 ≥ cos(A-B) from (f) above.
The only solution to the above 2 inequalities is that
1 = cos (A-B)!
Hence, A = B.
So, sin²AsinC + cos²A = 1.
sin ²AsinC = sin²A from (e) above.
From (c) above, sinA is not equal to 0. So, we get
sinC = 1 or C = 90 deg.
Since A = B and A+B+C = 180 deg., A = B = 45 deg.
So, C = 90 deg. and A = B = 45 deg. which is condition 2.
Hence condition (2) is necessary for condition (1).
Sufficiency:
sin 45 deg. = cos 45 deg. = 1/√2.
sin 90 deg. = 1.
So, if C = 90 deg. and A = B = 45 deg.,
sinAsinBsinC + cosAcosB = 1/2 + 1/2 = 1.
So, condition 2 is sufficient for condition 1.
I like this problem because it is a case where an equation is rather easily solved with the clever use of inequalities. (The solution is not my own but I remember it from years ago when a smart professor came up with it in India. Other solutions possibly exist but are probably harder.) |
