MHB Determine the type of triangle PQR

[anemone]

Determine all triangles $PQR$ for which $\cos P \cos Q+\sin P \sin Q \sin R=1$.

 

[kaliprasad]

Spoiler

because of symmetry in P and Q let us assume P > Q

$\sin\, R $ cannot be 1 as $\sin\, R =1$ means
$\cos\, P \cos \, Q + \sin\, P \sin\, Q = 1$
or $\cos(P-Q)= 1$
so $R = (P-Q)= \dfrac{\pi}{2}$
which is impossible
$\sin\, R $ cannot be $\lt1$ as $\sin\, R \lt 1$ means
$\cos\, P \cos \, Q + \sin\, P \sin\, Q \gt1$
or $\cos(P-Q)\gt 1$ which is impossible
so no triangle

Thanks to Opalg's comment below I revisited the ans and realized that I had done the mistake that
$\cos(P-Q)= 1$
=> $(P-Q) = \dfrac{\pi}{2}$
it should be
$P-Q = 0$ giving $P=Q=\dfrac{\pi}{4}$ and $R = \dfrac{\pi}{2}$

 

[Opalg]

Spoiler

because of symmetry in P and Q let us assume P > Q

$\sin\, R $ cannot be 1 as $\sin\, R =1$ means
$\cos\, P \cos \, Q + \sin\, P \sin\, Q = 1$
or $\cos(P-Q)= 1$
so $R = (P-Q)= \dfrac{\pi}{2}$
which is impossible

$\sin\, R $ cannot be $\lt1$ as $\sin\, R \lt 1$ means
$\cos\, P \cos \, Q + \sin\, P \sin\, Q \gt1$
or $\cos(P-Q)\gt 1$ which is impossible
so no triangle

[sp]What about $P = Q = \frac\pi4$, $R = \frac\pi2$?[/sp]

 

[anemone]

[sp]What about $P = Q = \frac\pi4$, $R = \frac\pi2$?[/sp]

Thanks Opalg for your comment!

Spoiler

because of symmetry in P and Q let us assume P > Q

$\sin\, R $ cannot be 1 as $\sin\, R =1$ means
$\cos\, P \cos \, Q + \sin\, P \sin\, Q = 1$
or $\cos(P-Q)= 1$
so $R = (P-Q)= \dfrac{\pi}{2}$
which is impossible
$\sin\, R $ cannot be $\lt1$ as $\sin\, R \lt 1$ means
$\cos\, P \cos \, Q + \sin\, P \sin\, Q \gt1$
or $\cos(P-Q)\gt 1$ which is impossible
so no triangle

Thanks to Opalg's comment below I revisited the ans and realized that I had done the mistake that
$\cos(P-Q)= 1$
=> $(P-Q) = \dfrac{\pi}{2}$
it should be
$P-Q = 0$ giving $P=Q=\dfrac{\pi}{4}$ and $R = \dfrac{\pi}{2}$

Thanks for participating, kali! Your answer is correct, i.e. $PQR$ is a right-angled isosceles triangle.

The other brilliant solution that I want to share here is shown as follow:

Spoiler

Common sense tells us that

$\cos (P-Q)\le 1$

$\cos (P-Q)=\cos P \cos Q+\sin P \sin Q \le 1$---(*)

But we are told that $\cos P \cos Q=1-\sin P \sin Q\sin R$

Hence inequality (*) becomes

$\cos (P-Q)=1-\sin P \sin Q\sin R+\sin P \sin Q \le 1$

$\cos (P-Q)=1+\sin P \sin Q(1-\sin R) \le 1$

We can conclude by now that $\sin R=1$ and $\cos (P-Q)=1$, which means $R=90^{\circ}$, $P=Q=45^{\circ}$, or $PQR$ is a right-angled isosceles triangle.