Geometry Problem

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

Click the figure below to see the problem 715 details.

## Friday, January 6, 2012

### Problem 715: Triangle, Altitudes, Parallel, Circumcircle, Angle, Measurement

Labels:
altitude,
angle,
circumcircle,
measurement,
orthocenter,
parallel,
triangle

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Angle B'A'C' = Pi - 2A = 62 deg

ReplyDeleteMy guess is x = 62 deg.

Yet to prove.

Circle A'B'C'(9-point circle)passes through H

Pravin, you seem to be 100% correct.

ReplyDelete/_BC'C=/_BB'C=90. Hence we conclude that B, B', C' & C lie on a circle whose diameter is BC. Now prove that H is the midpoint of BC and we're done for then HB=HC=HB' and hence /_HB'C=/_HCB'=/_C and /_C'B'A=/_A (for B,C,B' & C' are concyclic) and thus /_C'B'H=/_A=59

Now HC'=HB' if H is the centre of our new circle. Then /_HCB' also =59 and thus /_x=62.

All we now need to do is to prove that H bisects BC. Problem #714 must come into play in order to prove this.

Ajit

Proof of Problem 719 completed.

ReplyDeleteIn the usual notation,

A’D = A’C’ = R sin 2B

A’E = R sin 2C

A’F = R sin 2B.sin 2C/sin(B-C) = A’D sin 2C /sin (B – C)

F, D, H, E being concyclic,

A’F. A’H = A’D. A’E

Follows

So A’H = A’E sin(B - C) / sin 2C = R sin(B - C)

Now BH = A’B + A’H = c cos B + R sin(B - C)

= 2R sin C cos B + R sin(B – C)

=R(2 sin C cos B + sin (B – C)]

= R sin (B + C)

= R sin A = a/2 = BC/2

Hence H is the midpoint of BC.

Pravin,

ReplyDeletePlease explain how this comes about:

A’F = R sin 2B.sin 2C/sin(B-C) = A’D sin 2C /sin (B – C)

Ajit

It is known that angles of triangle A'B'C' are 180-2A,180-2B,180-2C resp'ly.

ReplyDeleteConsider triangle A'C'F

<A'C'F=180-<A'C'B'=2C

<A'FC'=<CFB'=<AB'F-<ACB = B - C

By Sine Rule

A'C'/sin(B-C) = A'F/sin 2C

Circle A'B'C' is the nine-point circle and its radius is known to be R/2

chord A'C' = 2.(R/2).sin <A'B'C'= R sin (180-2B)= R sin 2B etc

Awaiting for a shorter proof of Problem 715