Geometry Problem

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

Click the figure below to see the complete problem 861.

## Thursday, February 28, 2013

### Problem 861: Triangle, Incircle, Parallel, Tangent, Metric Relations

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http://img16.imageshack.us/img16/3367/problem861.png

ReplyDeleteLet r= radius of incircle and L is tangent point of GH to incircle

Connect OG,OA,OE,OH,OF,OC

Note that triangles AOG and HOG are right triangles and OE and OF are altitudes.

Relation in right triangle AOG give OE^2=r^2= EG.AE= 7

Relation in right triangle HOC give OF^2=r^2=GH.HC=2.x=7

So x=7/2

Let is OE=OF=r

ReplyDeleteangle DEM= angle DFM=90 degrees , why MD is diameter of circle

ED is perpendicular of AO and EM is perpendicular of GO. So, angle GOA=90 degrees.

Similarly , angle HOC=90 degrees.

From the right triangle GOA we have, OE^2=AE.EG=7 or, r^2=7 (1)

From the right triangle COH we have, OF^2=CF.FH=7 o r, r^2=2x (2)

From (1),(2) ,3x=7 , x=7/2

http://img268.imageshack.us/img268/5052/74920469.png

Let BG=y, BH=z.

ReplyDeleteSince BG/BA = BH/BC = GH/AC and BE = BF

y/(y+8) = z/(z+2+x) = 3/(7+x) and y+1 = z+2

xy+yz+2y = yz+8z

3x+3z+6 = xz+7z

xy+7y = 3y+24

y+1 = z+2

Solving those equations, we have

x=7/2, y=16/5, z=11/5

Proiectam G si H pe AC si daca folosim Teorema lui Pitagora obtinem (x+2)(x+2)-(x-2)(x-2)=8.8-6.6,de unde x=7/2

ReplyDeleteLet BG = p so that BH = p-1 since BE = BF

ReplyDeleteFrom similar triangles

p/(p+8) = (p-1)/(p+1+x) = 3/((7+x)

Each ratio = 1/(7-x) where I have used the algebraic identity that if a/b = c/d then each ratio = (a-c) /(b-d)

So 3(7-x) = 7+x and so x = 7/2

Sumith Peiris

Moratuwa

Sri Lanka