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.

 

[CellCoree]

Based on the given equation, we can determine that triangle PQR is a right triangle. This is because the equation $\cos P \cos Q+\sin P \sin Q \sin R=1$ is equivalent to the Pythagorean identity $\cos^2 P + \sin^2 P = 1$. Therefore, any triangle that satisfies this equation must have one angle that measures 90 degrees.

In addition to being a right triangle, there are also other triangles that can satisfy this equation. These include isosceles right triangles, where two angles measure 45 degrees and one angle measures 90 degrees, as well as scalene triangles where one angle measures 90 degrees and the other two angles are complementary.

Overall, the given equation $\cos P \cos Q+\sin P \sin Q \sin R=1$ has infinite solutions for triangles PQR, as long as one angle measures 90 degrees and the other two angles are complementary. This shows the versatility and applicability of the Pythagorean identity in various geometric contexts.