DG must be twice of inradius . Now trig bash like your life depends on it.

Geo ==> I won't do

From the OMC RMO livesolve. solved with rawlat_vanak, AdhityaMV, zvfrozel, and mathscrazy. We setup a 1-dimensional coordinate system with $A = 0$ and $D = 1$. We may compute $G = 2/3$. Assuming $O = k$ and $H = l$, we obtain that $2k+l = 2$ by using Euler Line. From the other condition, we get \[2-2k = 23-23l\]Solving the linear system, we obtain $k = 25/48$ and $l = 23/24$. Define $L$ as the midpoint of arc $BC$ not containing $A$. We see that $L = 2k = 25/24$. Let $I$ have coordinate $x \in (0, 1)$. Then we have \[(l-x)^2 = LI^2 = LD \cdot LA = \left(\frac{25}{24}-1\right)\left(\frac{25}{24}-0\right)\]which implies that $l-x = \pm 5/24$ and thus $x = 5/6$ (the other root must be ignored). As a result, we may check now that $G$ is the reflection of $D$ over $I$, and thus $G$ lies on the incircle of $\triangle ABC$. $\square$

Solved with SatisfiedMagma. Solution: Toss the system on to the cartesian plane with $O$ as the origin and $AD$ as the $y-$axis. It's well known $HG : GO = 2:1$. So, by using the ratio in the problem statement, we can deduce the following coordinates. Keep in mind that all the triangle centers $O,H,G$ and the incenter as well lie on the $A-$altitude. $A = (0,25)$ $O = (0,0)$ $G = (0,-7)$ $H = (0,-21)$ $D = (0,-23)$ Let $I$ be the incenter of $\triangle ABC$ located on $(0,y)$. By Euler's Theorem, we know that $OI^2 = R(R-2r)$. By putting this in the language of coordinates, we get that \[y^2 = 25^2 - 2\cdot 25 \cdot(y + 23) \implies y \in \{-15,-35\}\]Now, note that $ y \ne -35$, otherwise $I$ will lie outside the triangle which is absurd. Therefore $y = -15$. Finally to show that $G$ lies on the incircle, it is enough to show that $G$ is the antipode of $D$ on the incircle of $\triangle ABC$. This can be readily verified by the coordinates found above. Thus we're done. $\blacksquare$

Wow $23$ is so amazing Assume $HD = 2x$ and $OD = 23x$. Now by well know results $AH = 2OD$ and $OG:GH = 1:2$ we calculate other length in terms of $x$. Let $L = AH \cap (ABC)$ now as reflection of $H$ above $BC$ lie on $(ABC)$ we have $LD = 2x$. By using power of point $$LB^2=LD \cdot LA$$we get $LB = 10x$ so if $I$ is incenter, we get $ID = 8x$. But $GD = GH + HD = 14x+2x = 16x$. Hence $G$ lie on incenter.

Let $\angle A = 2x.$ Hence $\angle B$ and $\angle C$ are equal to $90 -x$. It is easy to prove that $HD$ = $2R.cosB.cosC$ = $2R.sin^2x$ and $OD$ = $R.cosA$ = $R.cos2x$ As, $23HD$ = $2OD$, we can see $23 sin^2x$ = $cos2x$ Or $sin x = 1/5$ Now, Let $E$ be the intersection of the incircle of triangle $ABC$ with side $AB$. It is trivial that $I,G,O,H$ will all lie on the same line i.e. $AD$ Claim: $AG.AD$ = $AE^2$ Proof: It is equivalent of proving $AE/AD$ = root$2$/root$3$ Join $DE$, $\angle BED$ = $\angle BDE$ = $45 + x/2$ as $BE$ and $BD$ are tangents $\angle EDA$ = $45 - x/2$ Now $AE/AD$ = $sin(45 - x/2)/sin(45 + x/2)$ = $(cos x/2 - sin x/2)/(cos x/2 + sin x/2)$ = $(1 - sin x)/(cos x)$ = $4$/*$2$root$6$) = root$2$/root$3$ Hence, by Power of point $G$ lies on the incircle of triangle $ABC$

I look at 23 i skip

OG! Not even touched

EASIEST PROBLEM CLAIM cos A = 23/25 by OD = R cos A and HD = 2R cos B cos C(distance of orthocenter to side) = 2R cos^2(90-A/2) = R(1 − cos A). BY MULTIPLE ANGLE FORMULAE Now GD = AD/3 = (OA+OD)/3 = 16R/25 GI=GD/2=8R/25 AND OA=R OD=RCOSA=23R/25 and r = 4RSIN(A/2)SIN(B/2)SIN(C/2)=4RSIN (A/2)SIN^2(90-A/4) DOING SOME COMPUTATION OR USE r = AD/(1+ csc(A/2)) = 48R/(25×6) = 8R/25 thus GI=r=ID hence proved ALTERNATIVELY THIS COULD BE DONE BY ASSUMING D TO BE THE ORIGIN AND B AND C ON X AXIS AND A ON POSITIVE Y AXIS THEN JUST HAVE PATIENCE TO BASH LIKE ANYTHING

Straightforward trig, which I initially bashed, coz that was simpler.