I Trigonometry problem of interest

[hutchphd]

@hutchphd a super result. :) With a couple algebraic steps one gets $3x^2=11^2+2^2+(2)(11)$. I'll let you write it out, but if you don't get to it in a day or so, I could post the 3 equations and their rather straightforward solution.

Please feel free. I have had an exhausting week and seem disinclined to do anything (feeling each of my 73 yrs I guess.....ah, well. I wish LaTeX was not such a chore for me)

 

[Charles Link]

@hutchphd Here is the solution:

$(ac)(bd)=2x^2+(11)(2)$ from Ptolemy
$(ac)^2=4x^2-2^2$ from Pythagoras and Thales
$(bd)^2=4x^2-11^2$ from Pythagoras and Thales

This gives $(ac)^2(bd)^2=(4x^2-2^2)(4x^2-11^2)=(2x^2+(11)(2))^2$.
We get $16x^4-4x^2(2^2+11^2)+11^2 *2^2=4x^4+8(11)x^2+11^2*2^2$
Thereby $12x^4=4(2^2+11^2+(2)(11) )x^2$.

Finally $3 x^2=2^2+11^2+(2)(11)$, which agrees with our previous results from @kuruman of post 30 and @renormalize of post 32.

Doing the arithmetic, we get of course that $x=7$. :) (I almost forgot to include that part. It was more important in some ways to get agreement with posts 30 and 32. LOL )

 

[Charles Link]

Looking back to the other posts, I see @Gavran also used Ptolemy's theorem previously in post 43, but the post 52 method gets the answer of $x=7$ as well as the arithmetically simple result of posts 30 and 32 without the extra calculations of post 43.

 

[Gavran]

Looking back to the other posts, I see @Gavran also used Ptolemy's theorem previously in post 43, but the post 52 method gets the answer of $x=7$ as well as the arithmetically simple result of posts 30 and 32 without the extra calculations of post 43.

You are right.

There is one more approach which is based on the figure from the post #39 and it is simpler than the one I have provided in the post #43.

$\cos(\angle ADC)=a/(2R)$ from The definition of cosine in the context of a right-angled triangle
$\angle ADC+\angle ABC=180^\circ$ from The inscribed angle theorem
$a^2+(2R)^2-2\cdot a\cdot2R\cdot\cos(\angle ADC)=x^2+b^2-2\cdot x\cdot b\cdot\cos(\angle ABC)$ from The cosine theorem

Above three equations imply
$\cos(\angle ADC)=(a^2-x^2-b^2+(2R)^2)/(2\cdot(a\cdot2R+x\cdot b))=a/(2R)\implies x=7$.

 

[Charles Link]

@Gavran I presume in your solution that you plugged in $R=x$.

You then get $a^2-b^2+3x^2 =2a^2+ab$
so that $3x^2=a^2+b^2+ab$.

Very good. :) You have a simple solution also, and these last two algebraic steps that I included show that your method here is also in complete agreement with posts 30 and 32.

You might also include that when you used the cosine theorem that both are equal to $c^2$.

 

[Charles Link]

Looking back over things, I do think @daverusin in post 41 has something of considerable merit. With his method, which I figured out in posts 45 and 46, he was able to spot a solution that @renormalize missed in his post 32. The algebra of finding where the line with slope of $m$ that passes through the point $(-1,-1)$, (which lies on the rotated ellipse), and intersects the rotated ellipse at the other points isn't terribly difficult, and I think it would be a good exercise for some of those with good math skills to try, to see how you then get the $a,b,$ and $x$, basically in their entirety, from his results. So far I have been the only one to give his post 41 a "like", so I think many have probably overlooked his post or didn't try to work out the details to see what he did.

Edit: The rotated ellipse is $x^2+xy+y^2=3$. It may not even be well-known to many, but those who have good math skills should know you can do the transformation $x'=x \cos{\theta}+y \sin{\theta}$ and $y'=-x \sin{\theta}+y \cos{\theta}$, and know that with $\theta=\pi /4$ that the $xy$ term becomes $(x')^2/2-(y')^2/2$, with the $x^2+y^2=(x')^2+(y')^2$ staying the same with the transformation. Thereby an ellipse with $(x')^2/2+(y')^2/6=1$ is the form it takes in the coordinate system that has the axes rotated 45 degrees, which is now in the standard form for an ellipse.

 

[hutchphd]

You are right.

Yes indeed. Through laziness I did not really look at the attempt.......bad form on my part (apologies!)

 

[Charles Link]

@hutchphd It gets harder as we get older. I am 70 now. We sort of have an excuse. Too many of the younger ones though I see are relying too much on google and want instantaneous answers, and don't want to take the time to wrestle with math problems the way our generation did. :)

 

[hutchphd]

Although I slighted @Gavran by not mentioning his attempt, I don't think anyone had called out Ptolemy's Theorem by name. How often do you get to cite Ptolemy directly?? Who wouldn't want to add that in.... lotsa fun....and pedagogically an impressive theorem IMHO.

 

[Charles Link]

Although I slighted @Gavran by not mentioning his attempt, I don't think anyone had called out Ptolemy's Theorem by name. How often do you get to cite Ptolemy directly??

In post 43 @Gavran cited it on his very last line in parentheses. It is so easy to overlook something like that though, because this thread has a lot of posts, and it can also be difficult to go back to the previous page of posts. The solution we have though, written out in post 52, is a more direct application of Ptolemy's theorem, where the answer of $x=7$ appears in short order.

 

[Charles Link]

Going back to post 56, I now tried, just for the fun of it, to see if I could get another solution using @daverusin 's method of post 41. I tried $m=7/3$, and I very readily got $a=11$, $b=131$, and $x=79$.
I tested the solution to see if we have that $x^2=(a^2+ab+b^2)/3$ and we do indeed. :)

and one more I got easily is with $m=8/3$ that gives $a=2$, $b=167$, and $x=97$.

and $m=9/4$ gives $a=23$, $b=218$, and $x=133$.

and $m=6/5$ gives $a=74$, $b=107$, and $x=91$. The method clearly works.

See also posts 46 and 47 for some further detail of the method.
It should be noted that for any rational coordinates on the ellipse, the slope $m$ of the line connecting those coordinates with the point $(-1,-1)$ will also be rational. Once again, when the expression for $y$ from the equation of the line is put into the ellipse expression, and we solve the quadratic for $x$, we get two solutions, both rational with one of the solutions being $x=-1$.

It might be worth mentioning that the method @renormalize used in post 32 was to try every possible integer solution and see if they satisfy $x^2=(a^2+ab+b^2)/3$. With @daverusin 's method, we simply use all rational $m's$ in the range where the line intersects positive $x$ and $y$ on the ellipse, and we find that every $m$ we try works.

For $m=7/5$ we get $a=71$, $b=143$, and $x=109$.