On the hypotenuse $ BC $ of an isosceles right triangle $ ABC $ let $ M,N $ such that $ BM^2-MN^2+NC^2=0. $ Show that $ \angle MAN= 45^{\circ } . $ Cristinel Mortici
Problem
Source: Romanian JBMO TST 2000, day 1, p. 4
Tags: geometry
zuss77
16.08.2019 17:42
So let's form a right triangle $MNP$ from this condition, so that $P$ lays outside of $ABC$, $MP=BM$, $NP=NC$. Let bisector of $\angle P$ meet $MN$ at $L$. $ML/PM=NL/PN$, $\implies ML/BM=NL/NC$. So homothety at $L$ maps $\odot MNP$ to $\odot ABC$, and $PA$ - bisector of $\angle P$. If $J$ - $MN$-touchpoint of $P$-excircle of $\triangle MNP$, then $MJ+MP=NJ+NP$ $\implies J$ - midpoint of $BC$ and $A$ - $P$-excenter of $\triangle MNP$. So $\angle MAN = (\pi - \angle P)/2 = 45^\circ$.
GammaBetaAlpha
16.08.2019 17:48
Previous years Rmo question
TurtleKing123
28.05.2020 12:45
Nice problem!
Let $M'$ be a point on side $BC$ such that $\measuredangle NAM' =45^0$
Let x be $\measuredangle M'AB$ after that we make an angle chase.
By the law of sines in $\triangle AM'B$ and $\triangle AM'N$ we have
$$ AM'=\frac{M'B \cdot sin(45)}{sin(x)} $$$$ AM'=\frac{M'N \cdot cos(x)}{sin(45)}$$So $M'B=\frac{M'N \cdot cos(x) \cdot sin(x)}{sin^2(45)}$
Analougsly, after doing the same thing in $\triangle ACN$ and $\triangle ANM'$ we get
$NC=\frac{NM' \cdot sin(45+x) \cdot sin(45-x)}{sin^2(45)}$
Now we have to prove
$$(\frac{M'N \cdot cos(x) \cdot sin(x)}{sin^2(45)})^2+(\frac{M'N \cdot sin(45+x) \cdot sin(45-x)}{sin^2(45)}) ^2=NM'^2$$$\iff (sin(x) \cdot cos(x))^2 + (sin(45+x) \cdot^2 sin(45-x))^2=sin^4(45) | \cdot 4$
$\iff sin^2(2x) +cos^2(2x)=sin^4(45) \cdot 4$
$\iff 1=4 \cdot (\frac{1}{\sqrt{2}})^4$ $\to$ that's true.
So $BM'^2+NC^2=M'N^2$
But also $BM^2+NC^2=MN^2$
After getting $NC^2$ from both equations we get
$M'N^2-M'B^2=MN^2-MB^2$
If $M' \in (MB) \implies LHS>RHS$
If $M' \in (MC) \implies RHS>LHS$
So, $M=M'$
$\mathbb Q$. $\mathbb E$. $\mathbb D$