Triangle Side and Angle Relationship Problem #121 - July 21st, 2014

[anemone]

Let $x,\,y,\,z$ be the three sides of a triangle, and let $a,\,b,\,c$ represents the angles opposite them.

If $x^2+y^2=1989z^2$, find $\dfrac{\cot a+\cos b}{\cot c}$.

--------------------
Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Let $x,\,y,\,z$ be the three sides of a triangle, and let $a,\,b,\,c$ represents the angles opposite them.

If $x^2+y^2=1989z^2$, find $\dfrac{\cot a+\cos b}{\cot c}$.

--------------------
Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

Hi MHB,

I'm so glad that Opalg has wrote to me that has brought me to the attention that there might be a misprint in this week high school POTW and he was right. The $\cos b$ as stated in second term in the sum of the numerator should be a $\cot b$.

I'm terribly sorry that this kind of typo/mistake happened again.

The corrected version of the problem is shown as followed:

Let $x,\,y,\,z$ be the three sides of a triangle, and let $a,\,b,\,c$ represents the angles opposite them.

If $x^2+y^2=1989z^2$, find $\dfrac{\cot a+\cot b}{\cot c}$.

 

[anemone]

Congratulations to the following members for their correct solutions:

1. Opalg
2. lfdahl
3. MarkFL

I'll show the solutions from all of these three members because each of them adopted different approach to solve for the problem.(Yes)

Here is Opalg's solution:

Spoiler

By the cosine rule, $$(1) \qquad \cos a = \frac{y^2+z^2-x^2}{2yz}.$$ Divide through by $\sin a$: $$(2) \qquad \cot a = \frac{y^2+z^2-x^2}{2yz\sin a}.$$ Next, the area of the triangle is $$(3)\qquad \text{Area} = \tfrac12yz\sin a = \tfrac12xz\sin b = \tfrac12xy\sin c.$$ Therefore $$(4)\qquad yz\sin a = xz\sin b = xy\sin c.$$ Substitute (4) into (2): $$(5)\qquad \cot a = \frac{y^2+z^2-x^2}{2xy\sin c}.$$ In exactly the same way as steps (1) - (5), but starting with the cosine rule for $\cos b$, $$(6)\qquad \cot b = \frac{x^2+z^2-y^2}{2xy\sin c}.$$ Add (5) and (6): $$(7)\qquad \cot a + \cot b = \frac{z^2}{xy\sin c}.$$ Now go back to the cosine rule in the triangle once again, using the information that $x^2+y^2 = nz^2$ (writing $n$ for $1989$ for convenience): $$(8)\qquad z^2 = x^2 + y^2 - 2xy\cos c = nz^2 - 2xy\cos c.$$ Therefore $$(9)\qquad xy\cos c = \frac{(n-1)z^2}2.$$ It follows that $$(10)\qquad xy\sin c = \frac{xy\cos c}{\cot c} = \frac{(n-1)z^2}{2\cot c}.$$ Now substitute (10) into (7): $$(11)\qquad \cot a + \cot b = \frac{2z^2\cot c}{(n-1)z^2},$$ and finally $$(12)\qquad \frac{\cot a + \cot b}{\cot c} = \frac2{n-1} = \frac2{1988} = \frac1{994}.$$

Here is lfdahl's solution:

Spoiler

The law of cosines:
\[cos\: a= \frac{y^2+z^2-x^2}{2yz}\\\\ cos\: b=\frac{x^2+z^2-y^2}{2xz}\\\\ cos\: c=\frac{x^2+y^2-z^2}{2xy}\]
The law of sines:
\[\frac{x}{sin\: a}=\frac{y}{sin\: b}=\frac{z}{sin\:c}\\\\ z = y\: \frac{sin\:c}{sin\:b}=x\: \frac{sin\:c}{sin\:a}\]
Thus:
\[cos\:a = \frac{y^2+z^2-x^2}{2yz}=\frac{y^2+z^2-x^2}{2yx\:\frac{sin\:c}{sin\:a}}\Rightarrow cot\: a = \frac{y^2+z^2-x^2}{2xy\: sin\:c}\\\\ cos\:b = \frac{x^2+z^2-y^2}{2xz}=\frac{x^2+z^2-y^2}{2xy\:\frac{sin\:c}{sin\:b}}\Rightarrow cot\:b = \frac{x^2+z^2-y^2}{2xy\: sin\:c}\\\\ cos\:c = \frac{x^2+y^2-z^2}{2xy}\Rightarrow cot\:c = \frac{1988z^2}{2xy\:sin\:c}\]
- where I have used the identity: $x^2+y^2 = 1989z^2$.

\[\frac{cot\:a+cot\:b}{cot\:c}=\frac{\frac{y^2+z^2-x^2}{2xy\:sin\:c}+\frac{x^2+z^2-y^2}{2xy\:sin\:c}}{\frac{1988z^2}{2xy\:sin\:c}}=\frac{y^2+z^2-x^2+x^2+z^2-y^2}{1988z^2}\\\\= \frac{2z^2}{1988z^2}=\frac{1}{994}.\]

And here is MarkFL's solution:

Spoiler

Let's begin with:

\(\displaystyle \cot(a)+\cot(b)=\frac{\cos(a)}{\sin(a)}+\frac{\cos(b)}{\sin(b)}=\frac{\sin(a)\cos(b)+\cos(a)\sin(b)}{\sin(a)\sin(b)}\)

Using the angle sum identity for sine, we may now write:

\(\displaystyle \cot(a)+\cot(b)=\frac{\sin(a+b)}{\sin(a)\sin(b)}\)

Now, since we know:

\(\displaystyle a+b+c=\pi\implies a+b=\pi-c\)

and coupling this with the identity:

\(\displaystyle \sin(\pi-\theta)=\sin(\theta)\)

we find:

\(\displaystyle \cot(a)+\cot(b)=\frac{\sin(c)}{\sin(a)\sin(b)}\)

Using the Law of Cosines, we may state:

\(\displaystyle z^2=x^2+y^2-2xy\cos(c)\)

We are given \(\displaystyle x^2+y^2=1989z^2\), hence:

\(\displaystyle z^2=1989z^2-2xy\cos(c)\)

\(\displaystyle xy\cos(c)=994z^2\)

\(\displaystyle \cot(c)=\frac{994z^2}{xy\sin(c)}\)

Now, we obtain:

\(\displaystyle \frac{\cot(a)+\cot(b)}{\cot(c)}=\frac{\sin(c)}{\sin(a)\sin(b)}\cdot\frac{xy\sin(c)}{994z^2}\)

The Law of Sines gives us:

\(\displaystyle \sin(c)=\frac{z}{x}\sin(a)=\frac{z}{y}\sin(b)\)

and so we now have:

\(\displaystyle \frac{\cot(a)+\cot(b)}{\cot(c)}=\frac{\frac{z}{x}\sin(a)}{\sin(a)\sin(b)}\cdot\frac{xy\frac{z}{y}\sin(b)}{994z^2}\)

\(\displaystyle \frac{\cot(a)+\cot(b)}{\cot(c)}=\frac{z^2\sin(a)\sin(b)}{994z^2\sin(a)\sin(b)}\)

Thus:

\(\displaystyle \frac{\cot(a)+\cot(b)}{\cot(c)}=\frac{1}{994}\)