## Sunday, December 18, 2011

### Problem 705: Intersecting Circles, Diameter, Angles, Measurement

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

Click the figure below to see the complete problem 705.

1. Happy New Year to all - I've a suggestion for Peter, Pravin and others. We've in our diagram E as the midpoint of GH and similarly F is the midpoint of MN. Thus, AH=AG and AM=AN.
What I cannot prove is that AG=AM & in fact AG=AH=AM=AN. In other words, a circle with centre at A and radius = AN will pass thru N,G,M & H and thus /_X=36 since/_NHG=36.
Now there remains a small matter of proving AG=AM where your expertise is solicited.
Cheerio
Ajit

2. Continuing the reasoning of Ajit:
In right triangle AMD: AM^2 = AF * AD ---(1)
In right triangle AHC: AH^2 = AE * AC ---(2)
Angle(ADE)=ACE) = 90–58 = 32 degree.
So AF * AD = AE * AC ---(3)
By (1),(2)and(3) ---> AM = AH.
So AH = AG = AM = AN,
therefore the quadrilateral HMGN is cyclic.
Then angle (NHG)=(NMG)=36 degrees.
So x = 36 degrees.

3. Thanks Anonymous.

4. GX*HX = CX*FX = DX* EX = MX* NX

Hence H,M,G,N is concyclic

Hence angle XMG = angle XHN = 36

1. NOTE : X is the point where EG intersects MF