perhaps solving this with euclidian backdated geo rather take toil and moil to the solver but without reserve i must say this if he really solve this with euclidian tools then he posesses quite good tactics in this era,
euclid perhaps fails here
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the arrivals
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let ABCD a square with center O and let M,N be the midpoints of segments BO,CD respectively.prove that triangle AMN is isosceles and right angled.
perhaps solving this with euclidian backdated geo rather take toil and moil to the solver but without reserve i must say this if he really solve this with euclidian tools then he posesses quite good tactics in this era,
perhaps solving this with euclidian backdated geo rather take toil and moil to the solver but without reserve i must say this if he really solve this with euclidian tools then he posesses quite good tactics in this era,
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN
Re: euclid perhaps fails here
Don't mind but I just could not digest your idea.
I think "solving with Euclidean backdated geo" is not much of a toil, though the term needs to be addressed properly. However, I think you should clarify what you mean. The solution is not much of tough, just applying Pythagorean Thm thrice does it
I think "solving with Euclidean backdated geo" is not much of a toil, though the term needs to be addressed properly. However, I think you should clarify what you mean. The solution is not much of tough, just applying Pythagorean Thm thrice does it
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor
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the arrivals
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Re: euclid perhaps fails here
sir do me a favour which terms make you feel confusion in solving the geometry?
and i am really much eager to see your solution if you incline to post the solution.because it really proved to me a hard nut to crack
..well perhaps it is relative
the arrivals
and i am really much eager to see your solution if you incline to post the solution.because it really proved to me a hard nut to crack
..well perhaps it is relativethe arrivals
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN
Re: euclid perhaps fails here
Euclidean geometry may not solve every part of the problem; however, in every nice solutions it works as the core.
My solution uses spiral similarity. In case you don't know these facts you can read this note by IMO gold medalist Zhao: http://web.mit.edu/yufeiz/www/similarity.pdf
Solution: Consider a spiral similarity $\mathcal{S}$ with center $A$, angle $\theta=45^o$, ratio $r=\frac{AC}{AB}=\sqrt{2}$.
$\mathcal{S}$ maps $\triangle ADC \to \triangle AOB$ as a result $\triangle ANC \to \triangle AMB$. This same spiral So there exists a spiral similarity that maps $\triangle ANM \to \triangle ABC$. (refer to the note.)
Q.E.D.
My solution uses spiral similarity. In case you don't know these facts you can read this note by IMO gold medalist Zhao: http://web.mit.edu/yufeiz/www/similarity.pdf
Solution: Consider a spiral similarity $\mathcal{S}$ with center $A$, angle $\theta=45^o$, ratio $r=\frac{AC}{AB}=\sqrt{2}$.
$\mathcal{S}$ maps $\triangle ADC \to \triangle AOB$ as a result $\triangle ANC \to \triangle AMB$. This same spiral So there exists a spiral similarity that maps $\triangle ANM \to \triangle ABC$. (refer to the note.)
Q.E.D.
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"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Re: euclid perhaps fails here
it can be done using only pythagorus's theorem and "segment joing midpoints of two sides of a triangle is half of the third side :-/
Let, $AB=a$
$\Longrightarrow AN^2 = \frac{5a^2}{4}$
and,
$AM^2 = AO^2 + OM^2 = {\frac{a}{\sqrt{2}}}^2 + {\frac{a}{2\sqrt{2}}}^2 = \frac{5a^2}{8}$
Let, $MZ || BC$ and $MZ$ intersect $CD$ on $Z$, and $AB$ on $Y$
clearly $NZ = \frac{a}{4}$, $MZ = a-MY = \frac{3a}{4}$
then, $MN^2 = MZ^2 + NZ^2 = \frac{5a^2}{8}$
So, $AM=MN$, and $AN^2 = AM^2 + MN^2$, Done!
Let, $AB=a$
$\Longrightarrow AN^2 = \frac{5a^2}{4}$
and,
$AM^2 = AO^2 + OM^2 = {\frac{a}{\sqrt{2}}}^2 + {\frac{a}{2\sqrt{2}}}^2 = \frac{5a^2}{8}$
Let, $MZ || BC$ and $MZ$ intersect $CD$ on $Z$, and $AB$ on $Y$
clearly $NZ = \frac{a}{4}$, $MZ = a-MY = \frac{3a}{4}$
then, $MN^2 = MZ^2 + NZ^2 = \frac{5a^2}{8}$
So, $AM=MN$, and $AN^2 = AM^2 + MN^2$, Done!
Last edited by Corei13 on Tue Jan 18, 2011 5:32 pm, edited 1 time in total.
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ishfaqhaque
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Re: euclid perhaps fails here
moon can't you give a note on spiral similarity? this note includes only problems"P
Re: euclid perhaps fails here
Oh...sorry. The correct link is http://web.mit.edu/yufeiz/www/three_geometry_lemmas.pdf
BTW you can find the rest of the awesome notes here: http://web.mit.edu/yufeiz/www/olympiad.html
BTW you can find the rest of the awesome notes here: http://web.mit.edu/yufeiz/www/olympiad.html
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
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the arrivals
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Re: euclid perhaps fails here
yes corei13 youe statement was claimed by avik roy earlier showing much efficiency over euclidean geometry.however nice solution.
transformational geometry is all greek to me
transformational geometry is all greek to me
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN