Is a,b fixed?

sanyalarnab wrote: Is a,b fixed? Yes.

Since $ABCD$ is a deltoid, $I$ lies on its axis of symmetry $AC$. Therefore, by bisector theorem on $BCA$, we have $AB:BC=AI:IC$, which implies $AI=\frac{a}{a+b}AC$. Since $C$ lies on a fixed circle of radius $b$ and center $B$, it follows that the locus of $I$ is this circle scaled with center $A$ and factor $\frac{a}{a+b}$ (which in fact passws trough $B$).

Well not exactly. First of all the problem says “non degenerate convex quadrilateral” but not all points on the circle satisfy this property.

@2 above since $ABCD$ is not degenrate, not all points on this circle work. My solution is same with at 2 above, but I find the locus as arc of this circle. Let circle with center $B$ and radius $b$ be $\Gamma$ and let Homothety with center $A$ and radius $\frac{a}{a+b}$ maps $\Gamma$ to $\omega$. Then it's obvious that $B$ lies on $\omega$. Let $AX_1$ and $AX_2$ tangent to $\omega$ at $X_1,X_2$, rescpectively. Then locus of $I$ is $\text{arc}$ $X_1BX_2$ of $\omega $, except points $X_1,X_2$ and $B$.

FishHeadTail wrote: Well not exactly. First of all the problem says “non degenerate convex quadrilateral” but not all points on the circle satisfy this property. Yes I omitted the fact that the two diametrically opposite on segment $AB$ don't work

By the way,I am curious about why the title is 2022 National Olympiad.Isn't that 2021?

Clearly $I$ lies on $AC$. Also, from the angle bisector theorem, $\frac{AI}{AC} = \frac{a}{a+b}$, so its obvious that the locus is the result of taking the locus of $C$ and performing a homothety $H\left(A, \frac{a}{a+b}\right)$ on it. As the locus of $C$ is clearly half of a circular arc, we are done.

Not a single complete solution... After obtaining $I$ is on that circular arc, as not all points must work, we need to still give some discussion. Then, we have to split into multiple cases, and I summarize the result: If $a=b$, the entire circle except two diametrically opposite points is the locus. If $a<b$, we can see there are two arcs with central angle $\cos^{-1} \frac ab$ If $a>b$, we can see there are two arcs with central angle $\cos^{-1} \frac ba$ I hope somebody can provide a good solution: without a complete discussion, one can only obtain 15 marks, which the committee correctly noted was too high, but this mark was given due to the fact that locus is not often tested. Indeed, the average mark is 18.6, for silver is >19, but still not 20 even for the training camp.