BdMO Online Forum

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,

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

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

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.

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!

moon can't you give a note on spiral similarity? this note includes only problems"P

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

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