Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 730 details.

## Wednesday, February 22, 2012

### Problem 730: Triangle, Altitude, Orthocenter, Perpendicular, Concurrent Lines

Labels:
altitude,
concurrent,
orthocenter,
perpendicular,
triangle

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http://img196.imageshack.us/img196/7341/problem730.png

ReplyDeleteFor clarity see sketch on the left of above picture

We will use Ceva’s theorem in this case.

Quadrilateral AA3A1A2 is cyclic with diameter AA1

We have ∠ (AA3A2)= ∠ (AA1A2)= ∠ (ABA1)= ∠B ( angles face the same arc AA2)

Similarly ∠ (AA2A3)= ∠AA1A3)= ∠ (ACA1)= ∠C

So sin(A4AA3)/sin(A4AA2)=cos(AA3A2)/cos(AA2A3)=cos(B)/cos(C) (angles complement) ----(1)

Similarly sin(B4BB3)/in(B4BB2)=cos(C )/cos(A) ----- (2)

And sin(C4CC3)/sin(C4CC2)=cos(A)/cos(B) -----(3)

Multiply expressions (1) x(2) x(3) we get

RHS= cos(B)/cos(C) x cos(C )/cos(A) x cos(A)/cos(B) =1

So Cevians AA4, BB4 and CC4 will concurrent per Ceva’s theorem

Join C₁A₁,A₁B₁,B₁C₁.

ReplyDelete∆BC₁A₁~∆BCA with ∠BA₁C₁=A=∠B₃B₂B,∠BC₁A₁=C=∠B₂B₃B

So C₁A₁∥B₃B₂. Given B₂B₃⊥BB4. Thus C₁A₁⊥BB4.

Let C₁A₁^BB4 = Y, A₁B₁^CC4 = Z, B₁C₁^AA4 = X

(^ denotes intersection)

BB4 divides C₁A₁ (at Y) in the ratio C₁X:XA₁

=BXcotC:BXcotA = cotC:cotA.

Similarly

CC4 divides A₁B₁(at Z)in the ratio cotA:cotB,

AA4 divides B₁C₁(at X)in the ratio cotB:cotC.

Product of the above three ratios being +1, it follows by Ceva,

AA4X, BB4Y, CC4Z are concurrent.

To Pravin:

ReplyDeleteRefer to your solution .

"Product of the above three ratios being +1" is not applicable to Ceva's theorem in this case.

please clarify.

Thanks Peter. I have recast my proof as follows.

ReplyDeleteLet BB4 be extended to meet AC at Y’

(similarly CC4 to Z’ on AB, AA4 to X’ on BC)

BB4 ⊥ B₃B₂

So ∠ABY’=∠B₃BB4=90°-∠B₂B₃B=90°-C and similarly ∠CBY’=90°-A

AY’:Y’C=(ABY’)/(BY’C)where (...)denotes area.

= [AB.BY’.sin∠ABY’]: [BC.BY’.sin∠CBY’]

= (AB:BC)(cosC:cosA)

= (sinC:sinA)(cosC:cosA)

= sin2C:sin2A

In the same way, CX’:X’B = sin2B:sin2C and BZ’: Z’A = sin2A:sin2B

So(AY’:Y’C).(BZ’:Z’A).(CX’:X’B) = 1

Hence by Ceva’s theorem applied to ∆ABC

the cevians AA4X’, BB4Y’, CC4Z’ are concurrent.

Hint:It's cut at circumcenter

ReplyDelete