Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 742 details.
Wednesday, April 25, 2012
Problem 742: Scalene Triangle, Orthocenter, Centroid, Circumcenter, Circumradius, Midpoint, Distance, Square, Metric Relations.
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ReplyDeleteDenote Vec(XY)= vector XY and Proj(XY)= algebraic projection length of vector XY over line HO
1. OGH is Euler line of triangle ABC so
O,G,H are collinear and OG= 1/2HG=MG ……( Property of Euler line)
2. G is centroid of triangle ABC so Vec(GA)+Vec(GB)+Vec(GC)=Vec(0)
And Proj(GA)+Proj(GB)+Proj(GC)=0 …..( property of Centroid)
3. GB is a median of triangle OBM => MB^2-BO^2= b^2-R^2=2.MO.Proj(GB)
Similarly with triangle OMC we have MC^2-CO^2=c^2-R^2=2.MO.Proj(GC)
And with Triangle OMA we have MA^2-AO^2=a^2-R^2=2.MO.Proj(GA)
Add each side of above 3 lines b^2+c^2+a^2-3.R^2=2MO.( Proj(GA)+Proj(GB)+Proj(GC))= 0
So a^2+b^2+c^2=3R^2
Let E be the midpoint of AC
ReplyDelete∆BGH~∆EGO
=> BH/EO = BG/GE = 2 (Median Property)
=> BH = 2EO = 2R cos B (Note OA = R and ∠AOE = (1/2)∠AOC =B)
=> BH²=4R²cos²B=4R²(1−sin²B)=4R²−AC² (note AC/sin B = 2R)
Next BG²=[(2/3)BE]²=(4/9)BE²=(2/9)(2BE²)
But AB²+BC²=2[BE²+EC²]=2BE²+(AC²/2) (Note EC=(1/2)AC
So 2BE²=AB²+BC²-(AC²/2) and
BG²= (2/9) [AB²+BC²−(AC ²/2)]
BM is a median of ∆BGH
So BH²+BG²=2BM²+2MG²
Now MG =(1/2)HG = (1/2)(2/3)HO (note G trisects HO as 2:1)
= (1/3)HO
So MG² = (1/9)HO² = (1/9)(9R²−AB²−BC²−AC²) using a well-known result
Thus 2b² = 2BM²= BH² + BG² − 2MG²
= 4R²−AC²+(2/9)[AB²+BC²−(AC²/2)]−(2/9)(9R²−AB²−BC²−AC²)
= 2R²−[1+(1/9)−(2/9)]AC²+[(2/9)+(2/9)]BC²+[(2/9)+(2/9)]AB²
=2R²−(8/9)AC²+(4/9)BC²+(4/9)AB²
So b²=R²−(2/9)(2AC²−BC²−AB²)
Similarly
c²=R²−(2/9)(2AB²−AC²−BC²) and
a²=R²−(2/9)(2BC²−AB²−AC²)
Hence a²+b²+c²=3R²