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I am trying to solve the problem of finding all triangles with angles $A$, $B$ and $C$ (in $[0,\pi]$) such that $\cos A\cos B+\sin A\sin B\sin C=1$.

In the case where the triangle has a right angle, it can be proved that $C=\frac{\pi}{2}$ (thus $C$ is the right angle and $A=B=\frac{\pi}{4}$. If we assume that the triangle is isosceles, then the triangle is flat or has a right angle (hence is the above solution).

My intuition says that these are the only solutions but I struggle to prove it. Any idea?

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marked as duplicate by lab bhattacharjee algebra-precalculus Nov 30 '14 at 16:32

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  • $\begingroup$ Have you tried using the prosthaphaeresis formula ? $\endgroup$ – Lucian Nov 30 '14 at 8:42
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    $\begingroup$ $A=B = \frac\pi 4, C = \frac \pi 2$ gives $LHS = 2$. $\endgroup$ – Nemo Nov 30 '14 at 8:43
  • $\begingroup$ @Nemo: I mistyped the equation. Sorry. $\endgroup$ – Taladris Nov 30 '14 at 8:56
  • $\begingroup$ @Michael You deleted your answer, but why? I think it is good. The second of your terms is nonnegative, so the only way that the total can be $1$ is if it equals $0$ (so $C=\pi/2$) and $\cos(A-B)=1$ (so $A=B-\pi/4$). This corresponds to the one solution OP has identified. $\endgroup$ – alex.jordan Nov 30 '14 at 9:15
  • $\begingroup$ For users who cannot see deleted answers, Michael has this answer: $\cos(A)\cos(B)+\sin(A)\sin(B)\sin(C)$ is the same as $\cos(A-B)-(1-\sin(C))\sin(A)\sin(B)$. Now $(1-\sin(C))\sin(A)\sin(B)$ is nonnegative, and $\cos(A-B)$ is at most $1$. It follows that for the total to equal $1$, $\cos(A-B)=1$ and $(1-\sin(C))\sin(A)\sin(B)=0$. So $A=B$, and $C=\pi/2$. And OP's intuition is correct. $\endgroup$ – alex.jordan Nov 30 '14 at 9:19