Let ABC be an acute triangle with AB=AC. Let D be the point on BC such that AD is perpendicular to BC. Let O,H,G be the circumcenter, orthocenter and centroid of triangle ABC respectively. Suppose that 2⋅OD=23⋅HD. Prove that G lies on the incircle of triangle ABC.
Problem
Source: RMO 2024 Q3
Tags: geometry, circumcircle, trigonometry
03.11.2024 14:25
DG must be twice of inradius . Now trig bash like your life depends on it.
03.11.2024 15:16
Geo ==> I won't do
03.11.2024 15:33
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=23−23lSolving 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∈(0,1). Then we have (l−x)2=LI2=LD⋅LA=(2524−1)(2524−0)which implies that l−x=±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 △ABC. ◻
03.11.2024 17:39
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 △ABC located on (0,y). By Euler's Theorem, we know that OI2=R(R−2r). By putting this in the language of coordinates, we get that y2=252−2⋅25⋅(y+23)⟹y∈{−15,−35}Now, note that y≠−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 △ABC. This can be readily verified by the coordinates found above. Thus we're done. ◼
03.11.2024 19:04
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∩(ABC) now as reflection of H above BC lie on (ABC) we have LD=2x. By using power of point LB2=LD⋅LAwe get LB=10x so if I is incenter, we get ID=8x. But GD=GH+HD=14x+2x=16x. Hence G lie on incenter.
03.11.2024 20:06
Let ∠A=2x. Hence ∠B and ∠C are equal to 90−x. It is easy to prove that HD = 2R.cosB.cosC = 2R.sin2x and OD = R.cosA = R.cos2x As, 23HD = 2OD, we can see 23sin2x = cos2x Or sinx=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 = AE2 Proof: It is equivalent of proving AE/AD = root2/root3 Join DE, ∠BED = ∠BDE = 45+x/2 as BE and BD are tangents ∠EDA = 45−x/2 Now AE/AD = sin(45−x/2)/sin(45+x/2) = (cosx/2−sinx/2)/(cosx/2+sinx/2) = (1−sinx)/(cosx) = 4/*2root6) = root2/root3 Hence, by Power of point G lies on the incircle of triangle ABC
05.11.2024 11:21
I look at 23 i skip
05.11.2024 13:39
OG! Not even touched
07.11.2024 10:33
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
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29.11.2024 18:51
Straightforward trig, which I initially bashed, coz that was simpler.