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## Saturday, February 21, 2015

### Geometry Problem 1088: Triangles, Perpendicular, Concurrent Lines

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concurrent,
perpendicular,
triangle

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From A₁A₂BB₃ concyclic, we have ∠C₁A₁O₁=∠A₂BB₃.

ReplyDeleteFrom A₁A₂CC₃ concyclic, we have ∠B₁A₁O₁=180°−∠A₂CC₃.

From B₁B₂AA₃ concyclic, we have ∠C₁B₁O₁=∠B₂AA₃.

From B₁B₂CC₃ concyclic, we have ∠A₁B₁O₁=∠B₂CC₃.

From C₁C₂AA₃ concyclic, we have ∠B₁C₁O₁=∠C₂AA₃.

From C₁C₂BB₃ concyclic, we have ∠A₁C₁O₁=180°−∠C₂BB₃.

Since A₁A₂, B₁B₂, C₁C₂ concurrent at O₁,

sin∠A₁B₁O₁ / sin∠C₁B₁O₁ × sin∠B₁C₁O₁ / sin∠A₁C₁O₁ × sin∠C₁A₁O₁ / sin∠B₁A₁O₁ = 1

Therefore,

sin∠B₂CC₃ / sin∠B₂AA₃ × sin∠C₂AA₃ / sin∠C₂BB₃ × sin∠A₂BB₃ / sin∠A₂CC₃ = 1

Hence, AA₃, BB₃, CC₃ concurrent at O₂.

http://s11.postimg.org/jzprw7z7n/pro_1088.png

ReplyDeleteLet a, a’, b, b’ and c, c’ are angles shown in the sketch

Let (UV, YW) represent angle form by rays UV and YW

We have (CB, Cz) = a …. ( A1, A2, C, C3 cocyclic)

And ( CA,Cz) =b ….. ( C, C3, B2 and B1 cocyclic)

Similarly (AC,Ax)=b’ and (Ax,AB)= c’

And (BA,By)= c and ( By,BC)= a’

Applying Ceva’s theorem in triangle A1B1C1 and rays A1A2, B1B2 and C1C2 we have .

(sin(a)/sin(a’)) x( sin(c)/sin(c’)) x ( sin(b’)/sin(b))= 1

In triangle ABC and rays Ax, By and Cz

Verifying that (sin(CB, Cz)/sin(CA, Cz)) x( sin(AC, Ax)/sin(Ax, AB)) x ( sin(BA, By)/sin(By, BC))=

(sin(a)/sin(a’)) x( sin(c)/sin(c’)) x ( sin(b’)/sin(b))= 1

So Ax, By and Cz are concurred at O2 per converse of Ceva’s theorem