## Thursday, November 25, 2010

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

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

1. Denote BC = 2y.
So 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,
In triangle ABD
sin <ABD = sin C = AD/AB,
1/sqrt3 = 2(sqrt2) / x
x = 2 sqrt (2/3)

2. The last step should read x = 2 sqrt6
Alternatively,
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

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

3. Problem 544

Draw 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

4. See link below for the sketch of the problem 544

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

Peter Tran

5. Draw EP//BD => EP = 2 => BD = 4 => BF = AF =3 =>
AD = 2√2 => x² = 4² + (2√2)²
x = 2√6

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)
But 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).

7. Problem 544
Fetch AM=//BE then B, F and M are collinear (<ACB=<ABD=α=<BAE ).Then AF=BF=FM=y.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

8. Draw EG // BD, G on AC

< 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