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Source: turkey TST 2007

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#1 h Apr 10, 2007, 12:37 PM • 2 Y

Y by Adventure10, Mango247

Two different points $A$ and $B$ and a circle $\omega$ that passes through $A$ and $B$ are given. $P$ is a variable point on $\omega$ (different from $A$ and $B$ ). $M$ is a point such that $MP$ is the bisector of the angle $\angle{APB}$ ( $M$ lies outside of $\omega$ ) and $MP=AP+BP$ . Find the geometrical locus of $M$ .

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#2 h Apr 10, 2007, 12:59 PM • 2 Y

Y by Adventure10, Mango247

Take point $K$ on half of arc $AB$ . We have that $KP=(AP+BP)\frac{AK}{AB}$ . So we see that locus of the points M is the arc (if point P is only on the one arc).

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#3 h Jun 12, 2007, 6:09 PM • 2 Y

Y by Adventure10, Mango247

why $KP=(AP+BP)\frac{AK}{AB}$

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#4 h Jun 13, 2007, 10:28 AM • 2 Y

Y by Adventure10, Mango247

According to the Ptolemy's Theorem $KP.AB=AK.PB+KB.AP\Rightarrow KP=(AP+PB).\frac{AK}{AB}=MP.\frac{AK}{AB}$ ok.
now, why locus of $M$ is an arc?

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#5 h Dec 12, 2012, 8:34 PM • 2 Y

Y by Adventure10, Mango247

Let $B'$ and $A'$ be on $AX$ and $BX$ beyond $X$ such that $AB'=BA'=AB$ and $X$ is the middle of arc $AB$ without $P$
locus of $M$ is the arc $A'B'$ on circle $A'B'X$ without $X$ .
hint:
i did it by using the centres $O_{1},O_{2}$ of the circles $PAB$ and $XA'B'$ and point $D$ which is the intersection of $PX$ and $AB$ . It's not so hard it's just some ratio chase that uses similar triangles and $AX=BX$ .

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#6 h Jan 26, 2013, 9:01 AM • 1 Y

Y by Adventure10

Not a different solution, but the complete result is a pair of arcs.

We have $AP+BP$ , angle bisector, a circumcircle. This is the special case of Ptolemy.
Let $N$ be the midpoint of $\overarc {AB}$ . By Ptolemy,
$PA\cdot BN + PB\cdot AN = PN\cdot AB$
$\Rightarrow (PA+PB)\cdot AN = PN \cdot AB$

$\Rightarrow \frac{PA+PB}{PN} = \frac {AB}{AN} = \text{Constant.}$

Let $A'$ be a point on the extension of the ray $[NA$ such that $AB=A'A$ . Since $A,N,B$ are constants, $A'$ is a constant point. We have

$\frac{MP}{PN}=\frac{AA'}{AN} = \frac {AB}{AN}$

So $A'M \parallel AP$ , and $\angle A'MN = \angle APN$ .
Similarly, construct $B'$ which is constant. We have $\angle A'MB' = \angle APB = \text{Constant}$ .
So $M$ is on the arc $%Error. "widearc" is a bad command. {A'B'}$ .
Since there is another $N$ , call it $N'$ , such that $N'$ is the midpoint of the arc $\overarc{AB}$ . Let those points be $A''$ and $B''$ .
The complete solution is that the loci are the arc $A'B'$ of $(A'NB')$ not containing $N$ and the arc $A''B''$ of $(A''N'B'')$ not containing $N'$ .

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#7 h Jun 7, 2013, 8:23 AM • 2 Y

Y by Adventure10, Mango247

Why, $M$ lies outside of $\omega$ ?
There are $A,B$ and $P$ such that $PA+PB<2R$ and $M(.)\in \omega$ .

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