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

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

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## Friday, July 5, 2019

### Geometry Problem 1441 Intersecting Circles, Perpendicular Bisector, Parallel Lines

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https://photos.app.goo.gl/VcpDVKKDqwxshkUg9

ReplyDeleteDefine points I, J, K, L, P, S,T, X and Y as per attached sketch

Note that I, J are midpoints of OD and CD

L and K are midpoints of OF and EF

QXJI and QYKL are rectangles

We have OJ= ½(OC+OD)= 1/2OC + OI

And JI= OJ-OI= 1/2OC= QX

Similarly, KL= ½ OE=1/2OC= QY

Since QX=QY => Chord GH= Chord MN=> Arc GH= Arc MN

And Arc GM=Arc NH => GN//MH

Let OS ⊥GN , we have quadrilaterals OSGJ and OSNK are cyclic

We have ∠ (COS)= ∠ (SGJ)= u=∠ (SNK)= ∠ (SOP)

So ∠ (COS)=u=∠ (SOE) => OS is the angle bisector of ∠ (TOP) and ∠ (COE)=> CE//GN

From previous problem, A,C,H co line, M,E,B co line.

ReplyDeleteConnect ACH, connect BEM.

Connect AB;

so angle AHM = angle ABM;

also angle CEM = angle ABE (ACEM on the same circle O),

so CE//MH

Let GN and OF cross at P;

OA=OB, so curve OA=curve OB;

curve AG= curve GD; curve BN=curve NF; so angle(OA)+angle(OB)+angel(AG)+angle(GD)+angle(DMHF)+angle(NF)+angle(BN)=180;

so angle NPF = angle(NF)+angle(OA)+angle(AG) = 90-1/2 angle(DMHF) =90- 1/2angleFOD = angle CEO.

So CE//GN;

So CE//GN//MH

If Point's

ReplyDeleteQ(R,0) and O(0,0)

Define:

(x-r)²+y²=R²

x²+y²=r²

On Point O, can define two lines

y=a*x (Line on Point's OCD) and y=b*x (Line on Point's OEF)

So:

(x-R)²+y²=R²

y=a*x

=> Point D [R/(1+a²),(a*R/(1+a²))]

(x-R)²+y²=R²

y=b*x

=> Point F [R/(1+b²),(b*R/(1+b²))]

x²+y²=R²

y=a*x

=> Point C [r²/(1+a²),(a*r²/(1+a²))]

x²+y²=R²

y=b*x

=> Point B [r²/(1+b²),(b*r²/(1+b²))]

Solve for the Slope k (germ. Steigung) at the Line on the Point's CE

k(1)=a-b (i)

On Point's CD can construction a Normal line, who will Intersect the (Big) Circle (-R,0)

n: y=-(1/a)*(x-(R+r²)/(2*(1+a²)))+a*(R+r²)/(2(1+a²))

=> The Intersect Point's:

G [(p+q)/(2(1+a²)),(a*(r²-R)-p/a)/(2(1+a²))] and H [(q-p)/(2(1+a²)),(a*(r²-R)+p/a)/(2(1+a²))]

The same procedure can be define the Point's M and N.

...

=> The Intersect Point's:

M [(u+v)/(2(1+b²)),(b*(r²-R)-u/b)/(2(1+b²))] and N [(v-u)/(2(1+b²)),(b*(r²-R)+u/b)/(2(1+b²))]

In wich:

p=a*√(R²(4a²+3)+r²(2R-r²)) , q=2a²R+R+r²

u=b*√(R²(4b²+3)+r²(2R-r²)) , v=2b²R+R+r²

__________

Now it can define a new Lines with Point's G and N, M and H.

The Slope from the new Lines, a solve will be show:

k(2)=a-b, line with Point's G and N (ii)

k(3)=a-b, line with Point's M and H (iii)

__________________________

So:

For the three Lines, the Slope is a same.

A prove for the Parallelity lines.

We use the results of problem n°1440 applied to both lines (O,C,D) and (O,E,F), and consider alignements (A,C,H) and (B,E,M) to which we apply Reim’s theorem (see problem n°72).

ReplyDeleteThus: MH//CE.

Similarly we obtain alignements (A,E,N) and (B,C,G) resulting in GN//CE.