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 $c^2$.