See complete Problem 103

Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Geometry Problem 103

Labels:
angle,
area,
equilateral,
heron,
interior point,
triangle

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ReplyDeleteHINT:Draw equilateral triangles CDE, ADF, and BDG. Apply SAS congruent triangles.where are points E,F,G?

ReplyDeletethanks

E, F, and G are exterior points in the plane of the triangle ABC so that CEAFBG is a hexagon.

ReplyDeleteSome errors have crept into my earlier proof. Please ignore the same. It must read as follows: ABC is the given equilateral triangle with a point D inside which is d, e & f units away from A, B & C resply. Rotate the triangle anti-clockwise thru' 60 deg. so that C goes to B, B to F and D to E. Now we've: AD=AE=d and BE=BD=e while FE=CD=f. It's easy to see that triangles DAC and EAB are congruent (corresponding sides equal) and, therefore, /_DAC = /_EAB which makes /_DAE=60 deg and since AD=AE, we conclude that triangle ADE is equilateral.

ReplyDeleteNow Tr. DAC + Tr. DAB = Tr. EAB + Tr. DAB = Tr. ADE + Tr. EDB and Tr. EDB has sides of d, e & f while Tr ADE is equilateral with side d.

Hence, Tr. DAC + Tr. DAB=(3)^(1/2)*(d^2)/4 +(s(s-a)(s-b)(s-c))^(1/2)/4 using Heron's Formula where 2s=d+e+f.

We can likewise prove that, Tr, DAB + Tr DBC=(3)^(1/2)*(e^2)/4 + (s(s-a)(s-b)(s-c))^(1/2)/4 and Tr. DBC + DCA =(3)^(1/2)*(f^2)/4+(s(s-a)(s-b)(s-c))^(1/2)/4

Add LHS and RHS of the three equations to get:

2*[Tr. ABC]=(3)^(1/2)*(d^2)/4+(3)^(1/2)*(e^2)/4 +(3)^(1/2)*(f^2)/4+3(s(s-a)(s-b)(s-c))^(1/2)/4 Or [Tr. ABC] = S =(3)^(1/2)(d^2)/8 +(3)^(1/2)*(e^2)/8+( 3)^(1/2)(f^2)/8+(3/8)( s(s-a)(s-b)(s-c))^(1/2) Or S=(1/2){(3)^(1/2)(d^2)/4 +(3)^(1/2)(e^2)/4+(3)^(1/2)(f^2)/4+3(s(s-a)(s-b)(s-c))^(1/2)} whics was to be proven.

Ajit

http://img198.imageshack.us/img198/538/problem103.png

ReplyDeleteDraw equilateral triangles CDE, DBG and ADF and we have a hexagon CEBGAF ( see picture)

We have triangles (CDB) congruence to (AGB) case SAS

similarly triangles (AFC) congruence to triangles (ADB) and triangles (CDA) congruence to triangles (CEB)

Area(CEBGAF)= 2.* Area(ABC)=2S

We also have congruence triangles EDB, GAD, CDF ( case SSS) with sides d, e, f .

Area(CEBGAF)= area(CDE)+Area(DBG)+Area(ADF)+3.Area(EDB)=2S

= f^2*SQRT(3)/4+ e^2*SQRT(3)/4+ d^2*SQRT(3)/4+3SQRT(s(s-e)(s-f)(s-d))

Divide both sides by 2 we will get the result

Peter Tran