A standard bash! Let the parabola be $y^2=4x$ with focus $(1,0)$ and let the points be $(t_1^2,2t_1),(t_2^2,2t_2),(t_3^2,2t_3)$. Obtain the following conditions: $$t_1t_2+t_2t_3+t_1t_3 = -5$$$$t_1+t_2+t_3+t_1t_2t_3=0$$Now use the determinant formula to find area of $ \Delta ABC$, use distance formula to find out semiperimeter, and divide the former by the latter to get the inradius as $2$.

@above can you explain furtherly what each condition represents and why?

Actually, we can say a bit more-- the incircle of $ABC$ remains unchanged, and $HI$ is the radius of the incircle as well as the symmetry axis of the parabola.

Dadgarnia wrote: Let points $A, B$ and $C$ lie on the parabola $\Delta$ such that the point $H$, orthocenter of triangle $ABC$, coincides with the focus of parabola $\Delta$. Prove that by changing the position of points $A, B$ and $C$ on $\Delta$ so that the orthocenter remain at $H$, inradius of triangle $ABC$ remains unchanged. Proposed by Mahdi Etesamifard

Supercali wrote: A standard bash! Let the parabola be $y^2=4x$ with focus $(1,0)$ and let the points be $(t_1^2,2t_1),(t_2^2,2t_2),(t_3^2,2t_3)$. Obtain the following conditions: $$t_1t_2+t_2t_3+t_1t_3 = -5$$$$t_1+t_2+t_3+t_1t_2t_3=0$$Now use the determinant formula to find area of $ \Delta ABC$, use distance formula to find out semiperimeter, and divide the former by the latter to get the inradius as $2$. I did try to solve something similiar but couldn't finish. Can you show that solution if you have it.

WLOG let the parabola be $y = x^2$. Let the points be $(a, a^2), (b, b^2), (c, c^2).$ The focus of the parabola is then $\left(0, \frac14\right).$ It can be shown that the incenter of the triangle is always $\left(0, \frac34\right).$ From here, we can select arbitrary $A, B, C$ satisfying the condition (e.g. $(0, 0), \left(\frac{\sqrt5}{2}, \frac54\right), \left(- \frac{\sqrt5}{2}, \frac54\right)$) to conclude that the incircle must have radius $\frac12$ (because the problem is true!), from which it would suffice to show that $H$ is on the incircle as in the above solution. So let's show that the incenter is always $\left(0, \frac34\right).$ First of all, the following will be necessary. Lemma. $ab + bc + ca = -\frac54$. Proof. We have from the condition of the problem that: $$\frac{c^2-a^2}{c-a} \cdot \frac{b^2 - \frac14}{b} = -1 \Leftrightarrow (c+a)\left(b^2 - \frac14\right) = -b.$$ This can be rearranged into: $$b^2c + b^2a + \frac54 b = \frac14(a+b+c).$$ Analogously, we get: $$c^2a+c^2b + \frac54c = \frac14(a+b+c).$$ Subtracting the two above equations gives: $$(c-b)\left(ab+bc+ca+\frac54\right) = 0.$$ Since $b \neq c$ is obvious, we have that $ab+bc+ca = -\frac54$ as desired. $\blacksquare$ We will show that $\left(0, \frac34\right)$ is on the $B-$internal angle-bisector of $\angle ABC.$ By the Tangent Addition Formula, we just require: $$\frac{2 \cdot \frac{b^2 - \frac34}{b}}{1- \left(\frac{b^2 - \frac34}{b}\right)^2} = \frac{a+b + c + b}{1 - (a+b)(c+b)}.$$ Cleaning this up a little, we want: $$\frac{2b\left(b^2 - \frac34\right)}{b^2 - \left(b^2 - \frac34\right)^2} = \frac{2b + (a+c)}{1 - b^2 - (ab+bc+ca)}.$$ Using $(a+c) = \frac{-b}{b^2 - \frac14}$ from the proof of the lemma and $ab+bc+ca = -\frac54$ from the lemma, we can rewrite the RHS as $$\frac{2b - \frac{b}{b^2 - \frac14}}{\frac94 - b^2} = \frac{2b^3 - \frac32b}{-\frac{9}{16} + \frac52 b^2 - b^4},$$ which means that we want: $$\frac{2b \left(b^2 - \frac34 \right)}{b^2 - \left(b^2 - \frac34\right)^2} = \frac{2b^3 - \frac32b}{-\frac{9}{16} + \frac52 b^2 - b^4}.$$ However this is obvious, since both the numerators and denominators match. So we've shown that $\left(0, \frac34\right)$ is the incenter of $\triangle ABC$. From here, finish as in the above solution. $\square$

MathDelicacy12 wrote: Dadgarnia wrote: Let points $A, B$ and $C$ lie on the parabola $\Delta$ such that the point $H$, orthocenter of triangle $ABC$, coincides with the focus of parabola $\Delta$. Prove that by changing the position of points $A, B$ and $C$ on $\Delta$ so that the orthocenter remain at $H$, inradius of triangle $ABC$ remains unchanged. Proposed by Mahdi Etesamifard

Where can we find official solutions?? Plzz provide the link as iam not able to find them on the official website

GammaBetaAlpha wrote: MathDelicacy12 wrote: Dadgarnia wrote: Let points $A, B$ and $C$ lie on the parabola $\Delta$ such that the point $H$, orthocenter of triangle $ABC$, coincides with the focus of parabola $\Delta$. Prove that by changing the position of points $A, B$ and $C$ on $\Delta$ so that the orthocenter remain at $H$, inradius of triangle $ABC$ remains unchanged. Proposed by Mahdi Etesamifard

Where can we find official solutions?? Plzz provide the link as iam not able to find them on the official website I attached the official solution of this problem and you can find the booklet here: https://igo-official.ir/events/

Similar Problem wrote: Let $H $ be the orthocenter of a $\triangle ABC $. Given that $H $ lies on the incircle of $(ABC)$. Prove that three circles with centers $A $,$B $, $C $ and radii $AH $, $BH $, $CH $ have a common tangent. Let $H_a $, $H_b $, $H_c $ be the feet of the altitudes. Note that $AH\cdot HH_a=BH\cdot HH_b=CH\cdot HH_c $. An inversion about a circle with center $H $, transforming $A $, $B $, $C $ to $H_a $, $H_b $, $H_c $ respectively. Iff the triangle is actue-angled takes a composition of the inversion and the reflection around $H $. This inversion transforms the sidelines of the triangle to the circles with diameters $AH$, $BH $, $CH $, and it transforns the incircle to the line touching the tree circles. A homothety with center $H $, and the coefficient $2$ transforms this line to the sought one.

SOLUTION USING MOVING POINTS: set the parabola $y=x^2$ and consider the circuconference $w$ of radius $1/2$ and center $(0;3/4)$ that will be the incircle s. t. if $ABC$ is tangent to it. by the triangle with vertices $(\sqrt5/2;5/4),(-\sqrt5/2;5/4),(0;0)$ and the ideal triangle with vertices in $(1/2;1/4),(-1/2;1/4)$ and the last vertex as the point of infinity of the parabola. Clearly becouse for the second triangle the triangle is tangent to the circle by porism this is always the case. It is also easy to check that the orthocenter of the two triangle is H in both cases: the first is evident, for the second the slope of the lines beetween the first and third point is $\sqrt5/2$ that is the same as minus one over the slope beetween the second and the focus and similarly for the other. Now i will prove that $\phi$ the transformation that sends a point A on the parabola to the line $r$ passing through the second point of the intersection of the tangents from A with tangency point Y to $w$ is projective(we will call B,C the second intersection of the tangents): Consider $\omega$ a projective transformation that sends the parabola and $w$ in 2 circle and denote with $*$ the transformed point/line. Then said M the midpoin of arc $B*C*$:$$A\rightarrow A*\rightarrow M\rightarrow Y* \rightarrow r*\rightarrow r$$is all projective, since from M to Y you can use $\delta$ the external homethety beetween the two circle as the tangent to the relative circle are clearly parallel. Now to conclude consider that if you parametrize A as a second degree point on the parabola the line AP (P the focus) is degree 2 so checking that $AP\perp r$ is of degree 4 so you just need to check 5 cases that are the 6 point I listed above. Remark: the transfromation $\omega$ may not be in $\mathbb{P}^2\mathbb{R}$ but if you work in $\mathbb{P}^2\mathbb{C}$ you can take the intersection of the curves to the circulars point at infinity

Here's a synthetic proof: Let $\ell$ be the directrix of $\Delta$ and $\tau$ be its axis of symmetry (i.e. $\tau$ is the line passing through $H$ perpendicular to $\ell$). Note that circles $$\odot(A,AH),\odot(B,BH),\odot(C,CH)$$are tangent to $\ell$. Let $\Phi$ denote the inversion at $H$ fixing $\odot(ABC)$. This sends $\odot(A,AH)$ to $BC$, etc. So $\Phi(\ell)$ must be the incircle of $\triangle ABC$. Let $\omega$ be the incircle of $\triangle ABC$ and $I$ be its center. We obtain $H \in \omega$ and $HI \perp \ell$. Next we show $\omega$ is fixed as $A,B,C$ vary. We use Poncelet's Porism. Let line through $H$ parallel to $\ell$ intersect $\Delta$ at points $X,Y$. By Poncelet's Porism, there exists a point $Z$ lying on $\Delta$ such that $ZX,ZY$ are tangent to $\omega$. Let $\mathcal R$ denote reflection in line $\tau$. Let $\mathcal R(Z) = Z'$. Clearly, $\mathcal R$ fixes $\omega$ and swaps $X,Y$. This means $Z'X,Z'Y$ are also tangents to $\omega$. This forces $Z \equiv Z'$. Hence $Z \in \tau$. Now since $\omega,\ell$ lies on the opposite sides of line $XY$, so this forces $Z$ to be the point at infinity along $\tau$, i.e. $Z$ is fixed. As $\omega$ touches $XY,ZX,ZY$, so $\omega$ is also fixed. This completes the proof. $\blacksquare$