## Saturday, November 1, 2014

### Geometry Problem 1054: Triangle, Altitude, Orthocenter, Circumcenter, Perpendicular, Midpoints

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it. 1. Obviously H is the orthocenter of ΔABC, and circle O is its nine-point circle.

Let P be the circimcenter of ΔABC, then BH=2×PF.
Now H,O,P are the Euler Line with HO=OP.
Also clearly, H,G,F are the median of ΔAHC, with HG=GF.

Thus, PF=2×OG, and hence BH=2×PF=4×OG.

2. http://s25.postimg.org/gymbfhbjz/pro_1054.png

Note that circumcircle of triangle MNF is 9 points circle of triangle ABC
Since N and F are midpoints of HC and AC => NF=1/2 AH=MH
And NF//MH => HNFM is a parallelogram
Since G is the midpoint of MN => H, G, F are collinear and GF=GH
Since ∠(LKF)= 90 degree => FOL is a diameter of circle O => GO= ½ GL= ¼ HB

3. Let BH intersect circle O (9-pt circle of ΔABC) at L.
First observe Δs LMN, FMN have same circumcircle.
L bisects BH. So LM // BA , so perp to HN. already BH perp MN.
So H is the orthocenter of ΔLMN.
Hence BH = 2 LH = 2(2 OG) = 4 OG..

4. Clearly O is center of 9 pt circle of ABC. Let P be mid-point of BH and Q foot of BH on AC. P,Q lie on the same circle.
Since m(PQF)=90=> P,O,F are collinear
PM // AB and m(PMF)=90 => MF // HN and similarly NF//MH => HMFN is parallelogram and G its centroid => HGF is one of its diagonal
Since G mid-point of HF and OG // PH => OG=PH/2=BH/4