Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Peter Tran, California, U.S.A.
Click the figure below to view more details of problem 1216.
Friday, May 20, 2016
Geometry Problem 1216: Circle, 90 Degrees, Tangent, Common Chord, Collinearity
Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Peter Tran, California, U.S.A.
Click the figure below to view more details of problem 1216.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Peter Tran, California, U.S.A.
Click the figure below to view more details of problem 1216.
Problem 1216
ReplyDeleteThe common tangents at points C and D intersect at E. The point E belongs to the radical axis of the circles with centers O1 and O2 as whereby the points A, B, E, are collinear.Is <ECO=
<EAO=<EDO=90,therefore the points O,A,C,E,D are concyclic.But CE^2=EB.EA=ED^2.So
Triangle CBE is similar with triangle ACE and triangle DBE is similar with triangle ADE.
Therefore <EBC=<ACE and <DBE=<ADE. But <DBE+<EBC=<ADE+<ACE=180. Therefore
The D,B,C are collinear.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS GREECE
Excellent solution.
DeleteSee below for the sketch:
http://s32.postimg.org/c1xovx2ut/pro_1216.png
Peter Tran
Пусть D' пересечение BC и окружность O, P основание высоты О1 до AB, <BCO1=a, и <ACO1=b.
ReplyDeleteO, O1, и C лежит на одной прямой, так как окружностей O и O1 прикоснется друг друга.
<PO1O=(<PO1A=a+b)-(<OO1A=2b)=a-b.
<O1OA=<PO1O=a-b потому что ОА и PО1 параллельны.
<OD'C=a,и <OAC=180-(a-b)-(b)=180-a, значит D', O, A, и C принадлежит одной окружность.
Тогда мы знаем <D'AC=<D'OC=180-2a.
<D'AB=<D'AC-<BAC=(180-2a)-(90-a)=90-a, значит описанная окружность D'AB касается окружность O в D', которые является самой точка D.