Geometry Problem

Click the figure below to see the complete problem 544 about Right Triangle, Altitude, Median, Equal angles, Measure.

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## Thursday, November 25, 2010

### Problem 544: Right Triangle, Altitude, Median, Equal angles, Measure

Labels:
altitude,
angle,
measurement,
median,
right triangle

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Denote BC = 2y.

ReplyDeleteSo BE = EC = y

From triangle ABC:

tan C = x/2y

In triangle ABE:

angle BAE = C and

tan C = y/x

Follows x^2 = 2 y^2,

tan C = y/x = 1/sqrt 2

Note angle ABD

= complement of angle DBC= C

Follows angle AFD = 2C

In triangle AFD:

cos 2C = 1/AF,

AF = [1 + (tan C)^2]/ [1 - (tan C)^2]

= [1 + (1/2)]/ [1 - (1/2)]

AF = 3

Also cos 2C = 1/3,

1 - 2 (sin C)^2 = 1/3,

sin C = 1/sqrt3

Now by Pythagoras

AD^2 = 9 - 1 = 8,

AD = 2 sqrt2

In triangle ABD

sin <ABD = sin C = AD/AB,

1/sqrt3 = 2(sqrt2) / x

x = 2 sqrt (2/3)

The last step should read x = 2 sqrt6

ReplyDeleteAlternatively,

ABE ~ CBA => x^2 = BC.BE = (1/2) BC^2

=> BC = x sqrt2 and BE = x/ sqrt2

AE^2 = x^2 + (x^2)/2 = 3x^2 /2, AE = x sqrt(3/2)

ADB ~ EBA => x^2 = AE.BD = [x sqrt(3/2)].BD

=> BD = x sqrt(2/3)

AD^2 = x^2 - BD^2 = x^2 - (2x^2/3) = x^2/3

AD = x/sqrt3

Finally AF^2 - AD^2 =1^2,

=> 3x^2/8 - x^2/3 = 1, x^2 = 24 , x = 2 sqrt6

Problem 544

ReplyDeleteDraw line NFM // BC ( N on AB , M on AC)

Let BE=EC= a

1. F is the midpoint of NM. (properties of // lines)

FN=FM=a/2

Triangle(ABE) similar to triangle (CBA) (case AA)

So BE/AB=AB/BC or a/x= x/(2.a) and x=a.SQRT(2)

2. AC=a.sqrt(6) and AM= a/2.sqrt(6) ( Pythagoras theorem)

3. Triangle (FDM) similar to triangle(ANM)

So FD/AN=FM/AM or 1/AN=a/(a.sqrt(6))

AN=sqrt(6) and AB=2.sqrt(6)

Peter Tran

See link below for the sketch of the problem 544

ReplyDeletehttp://img440.imageshack.us/img440/2296/problem544.png

Peter Tran

Draw EP//BD => EP = 2 => BD = 4 => BF = AF =3 =>

ReplyDeleteAD = 2√2 => x² = 4² + (2√2)²

x = 2√6

x^2=BE*(BC=2BE). F is circumcenter of AEB with radius R. <BDE=<FEB=90-a so BE^2=R(R+1).Then the original equation becomes x^2=2R(R+1)

ReplyDeleteBut from Pythagoras in triangle ABE x^2=(2R)^2-(BE^2=R(R+1))=3R^2-R

Now we have x^2=2R^2+2R=3R^2-R. Solving for R gets us R^2-3R=0

R is not 0 so with R=3, x^2=24, x=2sqrt(6).

Problem 544

ReplyDeleteFetch AM=//BE then B, F and M are collinear (<ACB=<ABD=α=<BAE ).Then AF=BF=FM=y.

Is triangle ADM similar triangle BDC so AD/DC=DM/DB=AM/BC=1/2.Therefore DC=2.AD,

BD=2.DM or y+1=2(y-1) or y=3,then AD^2=3^2-1=8, but AC=3.AD.

Then x^2=AB^2=AD.AC=3AD^2=3.8=24, or x=√24=2.√6.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Draw EG // BD, G on AC

ReplyDelete< ABD = á hence F is the centre of right triangle ABE. Let AF=FE=BF=y

From the mid point theorem, EG = 2 X DF = 4 hence y+1 = 2 X EG = 4 so y =3

Applying Pythagoras to right triangle ABE, x2 + a2 = 4y2 = 36…..(1)

Since AB is tangential at A to circle AEC, x2 = a2/2 …..(2)

From (1) and (2) ; 3x2 = 36

x = √24 = 2√6

Sumith Peiris

Moratuwa

Sri Lanka