Geometry Problem. Post your solution in the comments box below.

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

Click the figure below to see the complete problem 1019.

## Thursday, May 29, 2014

### Geometry Problem 1019: Circle, Diameter, Perpendicular, Chord, Tangent, Sum

Labels:
chord,
circle,
diameter,
perpendicular,
sum of segments,
tangent

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Algebra solution :

ReplyDeleteLet C’ is the point of sysmetry of C over AB => AC’=AC

Let α=∠(BAD) , AB= 2.R and AH= 2p

We have AD= 2R.cos(α) , DF=2R.cos(α) ^2 +2p

Since AD=DF => 2.R.cos(α)= 2p+2Rcos(α)^2 => R.cos(α)^2-R.cos(α)+p=0

Let α1 and α2 are solutions of this quadratic equation

Summation of solutions= 1 => cos(α1)+cos(α2)=1=> AD/AB+AC’/AB=1

Or AB=AD+AC

Consider:

ReplyDeleteR=DA, r = CA, 2a = AB(where O is the center)

Join B and D ->Triangle BDA is right triangle

Join B and C ->Triangle BCA is right triangle

Draw a radius DF (DF is perenpedicular to MN)

Draw a radius CE (CE is perenpedicular to MN)

Draw an tangent to circle O threw A in that way that it cut DF at L and CE at M thus MEFL is rectangle.

Triangle DAL is similiar to BDA and so MAC to BCA (because they has 90 angle of MEFL and becasue of the tangent : R/2a = DL/R (triangles DAL and BAD) - >DL = R^2/2a THUS LF = R - R^2/2a

AC/BA = MC/AC -> r/2a = MC/r (triangles CAM and BCA) - >MC = r^2/2a THUS ME = r - r^2/2a

Since MEFL is rectangle we get that ME = LF:

r - r^2/2a = R - R^2/2a

after short develop we get

2ar - r^2 = 2aR - R^2

R^2 - r^2 = (R - r)(R+r) = 2a(R - r)

thus 2a = R+r - > AB = DA + AC

Choose midpoint of AH to be origin and BH as y-axis.

ReplyDeleteIf a=AH/2 and r=AB/2, parabola 4ay=x^2 cuts circle x^2+(y-(a+r))^2=r^2 at C and D==>

4ay+y^2-2y(a+r)+(a+r)^2=r^2,

y^2-2y(r-a)+a^2+2ar=0,

CE+DF=(sum of y-coordinates of C and D)+2a=2(r-a)+2a=2r=AB

Let AC = a and AD = b and AB = c and AH = x

ReplyDeleteUsing Pythagoras

EH^2 = a^2-(a-x)^2 = a^2(c^2-a^2)/c^2 since EH.AB = BC.CA

So (a-x)^2 = a^4/c^2

Hence a-x = a^2 /c......(1)

Similarly b-x = b^2/c.....(2)

(1) - (2) a-b = (a^2-b^2)/c

Dividing both sides by a-b,

c = a+b

Sumith Peiris

Moratuwa

Sri Lanka

Note that E,A,D are collinear as well as F,A,C though this fact was not used in the above proof.

ReplyDeleteAnother problem from the above.,,,

ReplyDeleteProve that 1/AH = 1/AC + 1/AD

Also EH/ BC + FH/BD = 1

ReplyDelete