Let $A, B, C$ be a the angles of a triangle. Find the maximum value of $\sin A+\sin B+\sin C$ without calculus or Lagrange Multipliers.
My attempt: If we fix $A$, then $B+C$ is also fixed. We have $\sin B+ \sin C=2\sin\frac{B+C}{2}\cos\frac{B-C}{2}$. The maximum value of $\cos\frac{B-C}{2}$ is $1$ and it is attained when $B=C$. What should I do next?
$\begingroup$
$\endgroup$
5
-
$\begingroup$ Hint: Jensen’s inequality. $\endgroup$– Sungjin KimMay 7, 2019 at 5:13
-
$\begingroup$ math.stackexchange.com/questions/990418/… $\endgroup$– lab bhattacharjeeMay 7, 2019 at 5:17
-
$\begingroup$ I am not interested in a proof by Jensen's inequality because it is calculus related and I already know that proof. $\endgroup$– MathEnthusiastMay 7, 2019 at 5:29
-
$\begingroup$ @lab bhattaharjee thanks for the link, but it doesn't explain what to do after concluding that $B=C$ and this is where I am stuck. $\endgroup$– MathEnthusiastMay 7, 2019 at 5:30
-
1$\begingroup$ When $B = C$, $A=\pi - 2B$. Now rewrite the equation in terms of $B$. Find the condition on B and then find the minimum value. $\endgroup$– AbhinavMay 7, 2019 at 6:03
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
(1) $\sin A+\sin B+\sin C$ is the maximum when the area of the quadrilateral $PQRS$ is the maximum.
(2) With $A$ fixed, the area of $PQRS$ is the maximum when the area of $\triangle QRS$ is the maximum, i.e. when $R$ is the farthest point on the arc $QS$ from the chord $QS$. So, $R$ is the midpoint of arc $QS$. So, $B=C$.
(3) The area remains unchanged if $A$ and $B$ are interchanged. $PQRS$ is then an isosceles trapezium inscribed in the semi-circle.
(4) Reflect the trapezium in $PS$ to form a hexagon inscribed in the unit circle.
(5) The area of $PQRS$ is equal to the area of $QRST$. If it is maximized, $QRST$ should also be an isosceles trapezium. SO, $\angle RSP=\angle QOP$. Note that $\angle RSP=\angle QPS=\angle PQT$. Therefore, $\triangle OPQ$ is equilateral.
$\sin A+\sin B+\sin C$ is the maximum when $A=B=C=\dfrac{\pi}{3}$.
-
$\begingroup$ Wow, this is amazing! I hadn't thought about a geometric proof. Could you also please look at my work and tell me how to finish my idea? After getting that $B=C$ I got that $\sin A+\sin B+\sin C=2\sin B(1+\cos B) $ and I don't know how to maximise this. $\endgroup$ May 7, 2019 at 8:58
-
$\begingroup$ If $f(A,B,C)=\sin A+\sin B+\sin C$. $f(A,B,C)\le f(A,\frac{B+C}{2},\frac{B+C}{2})$. So $f(A,B,C)$ is not the maximum if $B\ne C$. Similarly, $f(A,B,C)$ is not the maximum if $A\ne B$ and $f(A,B,C)$ is not the maximum if $A\ne C$. If $f(A,B,C)$ has a maximum, it must happen when $A=B=C$. $\endgroup$– CY AriesMay 7, 2019 at 9:08