Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Omid Motahed.
Click on the figure below to view more details of problem 1262.
Thursday, September 22, 2016
Geometry Problem 1262: Isosceles Triangle, Altitude, Medians, 90 Degrees, Perpendicular
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Let the centroid be G.
ReplyDeleteSince BG:GH=2:1, thus AH=GH=CH.
Therefore H is the Orthocenter if AGC, with diameter AC.
Consequently, ∠AGC=90°
Typo: H is the circumcenter of AGC
DeletePalai Renato
ReplyDeleteBH/AC=1,5 we put AC=2 BH=3, BC= √10, CD= √10/2 , CD2= 2,5
The cross EC with AD =K; CK=X ; X2+X2=4, X=√2
ED=1 , KD=y ; y2+y2=1 y=1/√2
CD2= 2+1/2 = 2,5 therefore EC and AD are perpendicular
Reply
Let AD, CE meet at M and let AC = 2p
ReplyDeleteThen BH = 3p and MH = p
So AH = MH = CH = p
Hence < AMH = < CMH = 45 and so AD is perpendicular to CE
Sumith Peiris
Moratuwa
Sri Lanka
Problem 1262
ReplyDeleteBH is altitude and median then K is centroid of the triangle ABC with BH=3.KH. Then
3.KH=3.AC/2 or KH=AC/2 (AH=HC).Therefore AD is perpendicular CE.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
Comment sent by Sumith Peiris:
ReplyDeleteDear Omid and Antonio
Just realized...,
The triangle need not be isoceles for the medians AD and CE to be perpendicular.
It is sufficient that the median BH (as opposed to altitude) be 1.5 times AC
Hope u will adjust the problem accordingly.
Rgds
Sumith
Thanks Sumith
DeleteSee problem 1263 at:
http://www.gogeometry.com/school-college/3/p1263-triangle-medians-90-degree-perpendicular.htm
@Sumith, thanks that makes the problem even more interesting
ReplyDeleteextend AC to AZ such that HB=HZ
ReplyDeletethen LBZH=45.......so....LECH=45 etc.