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