Saturday, February 3, 2018

Geometry Problem 1353: Tangent Circles, Perpendicular Diameters, Congruence, 30 Degrees

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1353: Tangent Circles, Perpendicular Diameters, Congruence, 30 Degrees.

2 comments:

  1. https://photos.app.goo.gl/zWeKMpDAXjawGeye2

    Let 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

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  2. 1. Define ang(MXN)=2x & FB=s & BH=r & BD=d.
    2. 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.

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