Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 760 details.
Wednesday, June 6, 2012
Problem 760: Isosceles Triangle, Circle, Diameter, Altitude, Chord, Metric relations
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http://img705.imageshack.us/img705/9923/problem760.png
ReplyDeleteConnect lines per sketch.
Note that F and H are midpoints of AC and DC
In right triangle BFC we have CF^2=CB. CH= 1…. ( relation in right triangle)
In right triangle BEC we have CE^2=x^2=CD.CB=2CH.CB=2
So x=√ 2
x^2=CE^2=CD*CB & CD/CA=(CA/2)/CB or CA^2=2CD*CB or x^2=CA^2/2=(4/2)=2 hence x=√2
ReplyDeleteLet AC intersect circle at F.
ReplyDelete∠BFC is 90° (angle in semicircle).
So BF is ⊥ bisector of AC, and so CF = (1/2) AC = 1
Also A, B, D, E are concyclic (∵∠AFB = ∠ADB = 90°)
Now In the right ∆BEC, ED ⊥ BC.
So
x² = CE² = CD.CB = CF. CA = (1)(2) = 2, x = √2
I have two different solutions for this problem, but i dont have time to post it (video is comming soon). Anyways, my post have another intention, i want to comment that if triangle is obtuse, the problem is absurd. In my opinion is necessary tell that ABC is an acute triangle.
ReplyDeletehttp://www.mathematica.gr/forum/viewtopic.php?f=20&t=27359&p=133897
ReplyDeleteExtend CB to F such that CB = BF. Then < CAF = 90 and so 2^2 = CD.CF
ReplyDeleteSimilarly since < CAB = 90 x^2 = CD.CB
Comparing these 2 equations and since CF = 2. CB we have 2^2 = 4 = 2.x^2
So x = sqrt 2
Sumith Peiris
Moratuwa
Sri Lanka
Solution 2
ReplyDeleteLet AC cut the circle at F.
Since < BFC = 90, F is the mid point of AC.
Since AFDB is concyclic < CAD = < FBC = < FEC
Hence EC is a tangent to circle AEF at E and so x^2 = CF.CA = 1X 2 = 2
x = sqrt.2
Sumith Peiris
Moratuwa
Sri Lanka