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Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.

case 1- discriminant

We can rewrite the following equation

$ f(x) = x^2 - (p-1)x + p $

As we know the sum and product of $ \tan C $ and $ \tan B $

Settings discriminant greater than equal to zero.

$ { (p-1)}^2 - 4p \ge 0 $

This gives $ p \le 3 - 2\sqrt2 $. Or $ p \ge 3 + 2\sqrt2 $

solving both equation

$ A + B + C = \pi $

$ C + B + \frac{\pi}{4} = \pi $

$ C + B = \frac{3\pi}{4} $

Using this to solve both the equation give $ p \in $ real

I found this on Quora. https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle

the right method

$ 0 \lt B , C \lt \frac{3\pi}{4} $

Converting tan into sin and cos gives

$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $

Now using componendo and dividendo

$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $

We know $ \cos (B+C) = 1/\sqrt2 $

We know the range of $B$ and $C$ $(0, 3π/4)$ Thus the range of $B - C$. $(0, 3π/4 )$

Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$

Thus using this to find range gives $ P \lt 0 $ or $ p \ge 3+ 2\sqrt2 $

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7 Answers 7

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1) The second method is wrong because of a silly mistake.

2) The first method is wrong because apart from discriminant, it is also important to note that there are restrictions on the values of $ B $ and $ C $ , and thus their are restrictions on $ tan C $, and $ tan B $, and thus their are restrictions on $ p $.

When both $B$ and $C$ are acute angles, both the roots of the above equations are positive. thus $ p \gt 1 $.

When one of them is obtuse, $$ \tan B \tan C \lt 0 . $$

thus $$ p \lt 0 .$$

This with the intersection of non-negative discriminant, gives the correct answer.

Which give the range obtained in the third answer.

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I think that every approach should give correct solution. Different solutions appear if to do mistakes.

Conditions to $\angle B$ and $\angle C$ are WLOG \begin{cases} \angle B+\angle C = \dfrac{3\pi}4\\[4pt] \tan\angle B\tan\angle C = p\\[4pt] \angle B\in\left(0,\dfrac{3\pi}8\right]\\ \angle C\in\left[\dfrac{3\pi}8,\dfrac{3\pi}4\right).\tag1 \end{cases}

Also, are known relations

$$\tan(a+b) = \dfrac{\tan a + \tan b}{1-\tan a \tan b},\tag{2a}$$ $$\tan\dfrac34\pi = -1,\tag{2b}$$ $$\tan\dfrac38\pi = \frac{1+\cos\frac34\pi}{\sin\frac34\pi}=\sqrt2+1.\tag{2c}$$

Let $x=\tan \angle B$ and $y=\tan \angle C.$ then

$$y = \tan\left(\dfrac{3\pi}4-x\right) = \dfrac{-1-x}{1-x} = \dfrac{x+1}{x-1},\tag3$$ $$xy=p.$$

Therefore, $$P(p,x) = x^2+x-p(x-1) = 0.\tag4$$

If $\,\underline{x=y=\sqrt2+1}\,$ then $$P(p,\sqrt2+1) = 4+3\sqrt2 -p\sqrt2 = 0,\\$$ $$p=2\sqrt2+3\tag5.$$

If $\,\underline{x\in (0, \sqrt2+1)}\,$ then

$P(p,z)\,$ has not roots in the interval $\,z\in(-1,0)\,$ and has one root in the interval $\,z\in(0,\sqrt2+1),\,$ \begin{cases} P(p,-1)P(p,0) > 0\\[4pt] P(p,0)P(p,\sqrt2+1)<0, \end{cases}

\begin{cases} (p+1)p > 0\\[4pt] p(2\sqrt2+3-p)<0, \end{cases} $$p\in(-\infty,0)\cup(2\sqrt2+3,\infty).$$

Taking in account $(5),$ the answer is

$$\color{green}{\boxed{{\phantom{\Big|}\mathbf{p\in(-\infty,0)\cup[2\sqrt2+3,\infty)}.}}\tag6}$$

If the condition $(6)$ is satisfied, then the common solution is $$(\angle B,\angle C)\in\{(f_\angle(z_1),f_\angle(z_2)), (f_\angle(z_2),f_\angle(z_1))\},\tag7$$ where $$D=p^2-6p+1 = (p-3-2\sqrt2)(p-3+2\sqrt2),\tag8$$ $$z_{1,2} = \dfrac{p-1\pm\sqrt D}2,\tag9$$ $$f_\angle(z) = \arctan z +\dfrac\pi2(1-\operatorname{sgn} z).\tag{10}$$

Example 1. $\quad p=-\frac32,\quad z\in\{-3,\frac12\},\quad \angle B\approx 27^\circ,\quad \angle C \approx 109^\circ,\quad\angle B+\angle C = 135^\circ =\frac34\pi. $

Example 2. $\quad p=6,\quad z\in\{2,3\},\quad \angle B\approx63^\circ, \angle C \approx 72^\circ,\quad \angle B +\angle C = 135^\circ -\frac34\pi.$

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Let $C\leq B$ and $B+C=\frac{3\pi}{4}$.

i) When $\frac{3\pi}{8}\leq B<\frac{\pi}{2}$, then $\frac{\pi}{4}<C\leq \frac{3\pi}{8}$.

Define $$f=\tan\ B\tan\ ( \frac{3\pi}{4}-B)$$ The range of $f$ is $\{ \tan^2\frac{3\pi}{8} \leq t<\infty\}$, by (1) continuity, (2) considering $B\rightarrow \frac{\pi}{2},\ B\rightarrow \frac{3\pi}{8}$ and (3) \begin{align*} f' &=\frac{-\cos\ (2B-\frac{\pi}{4} ) }{\sqrt{2}\cos^2 B\sin^2 (B-\frac{\pi}{4}) } >0 \end{align*} when $\frac{3\pi}{8}<B$

ii) When $\frac{\pi}{2} < B<\frac{3\pi}{4}$, then $ -\infty<\tan\ B<T<0$ for some $T$. And $0<\tan\ C<1$ so that range of $f$ contains $\{-\infty < t<0\}$.

Hence note that these $B,\ C$ can be angles in a triangle so that $ \{ \tan^2\frac{3\pi}{8} \leq t<\infty\ {\rm or}\ -\infty<t<0\}$.

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For $A,B,C$ to be the angles of a triangle, not only it shall be $A+B+C = \pi$, but also $0 \le A,B,C$, or strictly greater than $0$ if you exclude the degenerate case.

So, the correct formulation of the problem is $$ \left\{ \matrix{ A = \pi /4 \hfill \cr \tan B\tan C = p \hfill \cr A + B + C = \pi \hfill \cr 0 \le A,B,C \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ \tan B\tan C = p \hfill \cr B + C = 3\pi /4 \hfill \cr 0 \le B,C\left( { \le 3\pi /4} \right) \hfill \cr} \right. $$

Taking care of this further restiction, whatever approach you follow (correctly) you will get to the unique right solution.

For instance, given $A=\pi /4$, we may start from the symmetrical case , isosceles triangle $B=C=3 \pi /8$ and put $$ \eqalign{ & \left\{ \matrix{ B = 3\pi /8 + D \hfill \cr C = 3\pi /8 - D \hfill \cr - 3\pi /8 \le D \le 3\pi /8 \hfill \cr \tan \left( {3\pi /8 + D} \right)\tan \left( {3\pi /8 - D} \right) = p \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad \left\{ \matrix{ B = 3\pi /8 + D \hfill \cr C = 3\pi /8 - D \hfill \cr - 3\pi /8 \le D \le 3\pi /8 \hfill \cr t = \tan \left( {3\pi /8} \right) = \sqrt 2 + 1 \hfill \cr - t \le x = \tan D \le t \hfill \cr {{t^2 - x^2 } \over {1 - t^2 x^2 }} = p \hfill \cr} \right. \cr & \Rightarrow \quad \left\{ \matrix{ t^2 = 3 + 2\sqrt 2 \hfill \cr 0 \le x^2 \le t^2 \hfill \cr p(x) = {{t^2 - x^2 } \over {1 - t^2 x^2 }} \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad p \in \left[ {t^2 = 3 + 2\sqrt 2 , + \infty } \right) \cup \left( { - \infty ,0} \right] \cr} $$

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The reason why the 1st method is wrong :

We can rewrite the following equation

$ f(x) = x^2 - (p-1)x + p $

As we know the sum and product of $ \tan A $ and $ \tan B $

Settings discriminant greater than equal to zero.

It seems that you meant "the sum and product of $\color{red}{\tan C}$ and $\tan B$".

Considering the condition that the discriminant is greater than or equal to zero is not enough because we also have to have $0\lt B\lt \frac 34\pi$.

This means that we have to consider the condition that $f(x)=0$ has at least one solution such that $x\lt -1$ or $x\gt 0$.

Therefore, the 1st method is wrong.


The reason why the 2nd method is wrong :

In the solution in Quora,

tan B + tan C = tanB tanC - 1

tanB+tan(3pi/4 - B) = p

This step is wrong. This should be $$\tan B+\tan\left(\frac 34\pi-B\right)=\color{red}{p-1}$$ Then, we get $$\tan^2B-(p-1)\tan B+p=0$$ But, again, similarly as the 1st method, considering the condition that the discriminant is greater than or equal to zero is not enough because we also have to have $0\lt B\lt\frac 34\pi$.

Therefore, the 2nd method is wrong.


To make the two methods correct, we have to consider the condition that $$x^2-(p-1)x+p=0$$ has at least one solution such that $x\lt -1$ or $x\gt 0$.

Solving this gives $$p\in (-\infty,0)\cup [3+2\sqrt 2,\infty)$$ which is the same as the answer in the third method.

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Let $\tan\beta=t$.

Thus, $t\neq1$, otherwise $\gamma=\frac{\pi}{2},$ which is impossible.

Also, $$0<\beta<\frac{3\pi}{4},$$ which is $$0<\beta<\frac{\pi}{4}$$ or $$\frac{\pi}{4}<\beta<\frac{\pi}{2}$$ or $$\frac{\pi}{2}<\beta<\frac{3\pi}{4},$$ which is $$0<t<1$$ or $$t>1$$ or $$t<-1.$$

Consider three cases.

  1. $t>1$.

By AM-GM we obtain: $$p=\tan\beta\tan\left(\frac{3\pi}{4}-\beta\right)=t\cdot\frac{-1-t}{1-t}=\frac{t^2+t}{t-1}=\frac{t^2+t-2+2}{t-1}=$$ $$=t+2+\frac{2}{t-1}=3+t-1+\frac{2}{t-1}\geq3+2\sqrt{(t-1)\cdot\frac{2}{t-1}}=3+2\sqrt2.$$ The equality occurs for $t-1=\frac{2}{t-1}$ and since $\lim\limits_{t\rightarrow1^+}p=+\infty$ and $p$ is a continuous function on $(1,+\infty)$, we got a range of $p$ in this case: $[3+2\sqrt2,+\infty).$

  1. $0<t<1$ We obtain: $$p=-\frac{t(1+t)}{1-t}<0.$$ Also, $$\lim\limits_{t\rightarrow1^-}p=-\infty,$$ $$\lim\limits_{t\rightarrow0^+}p=0$$ and $p$ is a continuous function on $(0,1),$ which says that the range of $p$ in this case it's $(-\infty,0).$

  2. $t<-1$.

In this case $p<0$, $$\lim_{t\rightarrow-1^-}p=0,$$ $$\lim_{t\rightarrow-\infty}p=-\infty$$ and $p$ is a continuous function on $(-\infty,-1).$

Thus, the range of $p$ in this case it's $(-\infty,0),$ which gives the answer: $$(-\infty,0)\cup[3+2\sqrt2,+\infty).$$

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We should consider discriminants but in the way they arise directly:

Let $tx = \tan(x)$ for shorthand notation

$$t_A= t{(\pi-B-C)}$$

$$1=\frac{-(tA+tB)}{1-tA\, tB} $$

$$tA \,tB=p\quad$$ plug in and simplify to find quadratic equation roots $$ tB^2+(1-p)tB+p=0 $$ $$ 2\,tA=-(1-p)-\sqrt{1-6p+p^2} $$ $$ 2\,tB=-(1-p)+\sqrt{1-6p+p^2}$$

For this to be real, quantity under radical sign should be positive.

The roots of quadratic equation then supply upper and lower limits automatically.

$$p_{lower}= 3+2\sqrt{2}\quad p_{lower}= 3-2\sqrt{2} $$

when $p$ should not lie between these limits as the quantity under radical sign is:

$$ \sqrt{p^2 -6p +1 } = \sqrt{(p-p_{lower})(p-p_{upper})} $$

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