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Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, February 3, 2018
Geometry Problem 1353: Tangent Circles, Perpendicular Diameters, Congruence, 30 Degrees
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https://photos.app.goo.gl/zWeKMpDAXjawGeye2
ReplyDeleteLet r and R are radius of circles A and B
Draw HP ⊥DH and NI ⊥ to DC ( see sketch)
In right triangle DHP we have DH^2=DB. DP
In right triangle DGC we have BG^2= DB. BC
So FG^2=4. BG^2= DB.(4BC)
Since FG=DH => DP=4.BC= 4.R
Since NC // HP So DN/DH= DC/DP=2r/(4R)=NI/BH=NI/R => NI=1/2.r
Triangle AIN is 30-60-90 degrees triangle.
So ∠ (MCN)=1/2 ∠ (MAN)= 30 degrees
1. Define ang(MXN)=2x & FB=s & BH=r & BD=d.
ReplyDelete2. CBNH is cyclical because ang(CBH)=ang(CNH)=90°.
3. Because of (2), ang(BHD)=x.
4. Define L as the midpoint of DH. We know that DL=LH=BG=FB=s.
5. Define Y on BL and perpendicular to D, let's call height YD=h.
6. Since tri(BYD)~tri(DBH), we have that h=r×d/(2s).
7. From chords FG and DC intersecting at B we have d=r×d/s².
8. From (6) and (7), we have that h=s/2.
9. ang(BLD)=2x because L is midpoint of DL and ang(DBL)=90°.
9. Triangle DYL is right and DL=YD/2. Then, 2x=30°. Q.E.D.