## Sunday, December 20, 2009

### Problem 405. Quadrilateral, 60, 75, and 135 degrees, Midpoint

Proposed Problem
Click the figure below to see the complete problem 405 about Quadrilateral, 60, 75, and 135 degrees, Midpoint.

Complete Problem 405
Level: High School, SAT Prep, College geometry

1. A slightly different way with one application of the sine rule gives us the following solution: Join A to C & E. Tr. ACD is equilateral since /_D=60 and CD=AD=2. Thus, /_CAD=60 & /_BAC=15,/_AEC=90, AE=√3 while /_ACE=60. Further, quad ABCE is concyclic & thus /_ABE = /_ACE=60 and finally tr. ABE gives, x/√3 = sin(45)/sin(60) resulting in x=√2
Ajit

2. join A to C, draw middle line EF = 1 ( F on AC ), join B to F. Win right isosceles tr BFE, BF = Ef = 1, angBFE=90
(BF is median for tr ABC)Pythagore theorem => x = v2

3. Purely Plane Geometry (No Need to Use Trigonometry):
Connect AC and AE, Extend Ray AE and Ray BC till they intersect (call the intersection F).
Since AD=CD=2 and Angle ADC=60 Degrees, ACD is equilateral. (AC=2, Angle ACD=60 Degrees.) AE=√3.
In ACE and ADE, AC=AD, Angle ACD=Angle ACD, CE=ED (given midpoint), so ACE congruent to ADE, Angle CAE=30 Degrees. Angle BAC=Angle BAD-Angle CAD=(75-60)Degrees=15 Degrees. Angle BAE=Angle BAC+Angle CAE=(15+30)Degrees=45 Degrees. So it's revealed ABCE is cyclic quadrilateral (Angle BCE+Angle BAE)=(135+45)Degrees=180 Degrees.
According to Ptolemy's Theorem, AC*BE=BC*AE+AB*CE. So we have 2x=√3*BC+AB. Angle BAE=Angle CFE=Angle FCE=45 Degrees. AF=√3+1. BF=AB=(√6+√2)/2, BC=(√6-√2)/2. Calculate:
2x=√3*(√6-√2)/2+(√6+√2)/2=2√2,
Therefore, x=√2.

4. EX the perpendicular from E to BC is 1/sqrt2, < ECX being 45. But EX = x/2 ence x = sqrt2.

Sumith Peiris
Moratuwa
Sri Lanka

5. Problem 405 - 2nd pure geometry solution.
Let AF be an altitude in triangle ABE.
Now ACD is equilateral.
In 30-60-90 triangle AED, AE = √3
In right isosceles triangle AFE, AF = FE = AE/√2 = √(3/2) …(1)
In 30-60-90 triangle BEF, FE = √3/2 x…(2)

From (1) and (2), √(3/2) = √3/2 x and so x = √2.

Sumith Peiris
Moratuwa
Sri Lanka

6. Prøblem 405 - 3rd pure geometry solution

Let AB and DC extended meet at P.
Tr. ACD is equilateral and Tr. PBC is right isosceles. Further ABCE is concyclic.

So Tr.s PAC and PEB are similar.

Hence x/2 = PB/PC = 1/sort 2

Therefore x = sqrt2

Sumith Peiris
Moratuwa
Sri Lanka

7. Prøblem 405 - 4th pure geometry solution

Let EQ
be the altitude of Tr. ABE.

Then EQ = sqrt3/2 x and since Tr. AQE is right isosceles this = sqrt(3/2)

x/2 = 1/sqrt2 and x = sqrt2

Sumith Peiris
Moratuwa
Sri Lanka

8. Prøblem 405 - 5th pure geometry solution.

Let AE, BC meet at R. Tr.s RBE and RAC are similar and Tr. CER is right isoceles.

So RE / CE = 1/sqrt2 = x/2 and hence x = sqrt2

Sumith Peiris
Moratuwa
Sri Lanka

9. See the drawing

DC=DA and ∠ADC= π/3 => ΔADC is equilateral
∠ACD= π/3, ∠BCA+∠ACD=∠BCD=155° => ∠BCA=75°
∠BAC=15° et ∠BCA=75° => ∠ABC= π/2

ΔADC is equilateral and EC=ED => ∠AEC= π/2
∠ABC= π/2 et ∠AEC= π/2 => A, B, C and E are concyclic with center O middle of AC => OA=OB=OC=OE=1
∠BAC=15° =>∠BOC=30° (Inscribed angle)
∠BOC=30° and ∠COE= π/3 => ∠BOE= π/2
OE=OB=1 and ∠BOE= π/2
Therefore BE=sqrt(2)

10. <ABC=90 (<sum of polygon)
Join AC
AC^2=2^2+2^2-2(2)(2)cos60
AC=2
So triangle ADC is an equilateral triangle
<ACD=60 & <ACB=75
cos75=BC/2
(sqrt6-sqrt2)/4=BC/2
BC=(sqrt6-sqrt2)/2

Consider triangle BCE and by cosine law
x^2=BC^2+1^2-2(BC)(1)cos135
x^2=2-sqrt3+1-(sqrt3-1)
x=sqrt2