If you can......?

Question

1)ABC is any triangle.Equilateral triangles ABR, BCP and AQC are described on the sides AB, BC, and AC respectively.
Prove that AP=BQ = CR.
(doesn't matter if you can't give diagram, keep the labelling same.)

2)If a, b, c are real numbers, all lying between 0 and 1, such that a+b+c= 2, show that

abc
________>=8
(1-a)(1-b)(1-c)

3)Determine all values of x if
0.5log(2x-1)+0.5log(x-9) >1
Log means log base 10.

Yes, it is from SMO.

Answer 1

2) Here you need to remember this :

(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac)

(a+b+c)^3 = a^3+b^3+c^3+3(a+b+c)*(ab+ac+bc) -3abc

Now you have :

abc >= 8[(1-a)(1-b)(1-c)]

you can do that, because, 1-a > 0, 1-b > 0 and 1-c > 0

then : abc / 8 >= (1 -b -a +ab)(1-c)

abc / 8 >= 1 - c - b +bc - a +ac +ab - abc

9abc / 8 >= 1 -a -b -c + ab + ac +bc

9abc / 8 >= ab + bc +ac -1

Now using the first properties I wrote :

4 = a^2 + b^2 + c^2 + 2(ab + bc + ac)

and

8 = a^3 + b^3 + c^3 + 6(ab + bc +ac) - 3abc

3) Here just do this :

0.5log((2x-1)*(x-9)) > 1


but there are some values of x that cannot be consider :

2x-1 > 0 and x-9 > 0

x > 1 / 2 and x > 9

0.5log(2x^2 - 18x - 1 + 9) > 1

2x^2 - 19x + 9 > 100

2x^2 -19x - 91 > 0

here this will be : ( 2x + 7)*(x - 13) > 0

the critical points are : -7 / 2 and 13

so using the linear graphic :

x E [ - infinite to -7/2 ] U [ 13 to infinite ]

BUT, x > 1 / 2 and x > 9

So then the only values for x, will be :

X E [ 13 to infinite ]

Hope that helped you, the first one is kind of difficult, is geometry, and I haven't done geometry exercises for a year

Answer 2

Looks as though the only answer you really need is no.1

In your diagram, draw lines RC and QB, and let them meet at Z.
Imagine a rotation about A of 60°: it will move triangle CAR onto triangle QAB - the eqilateral triangle sides ensure this. ∴ angle BZR = 60° and CR = QB. You can do the same again by rotating through 60° at B to show AP = CR, so AP = BQ = CR.

You have the other answers; I don't need to give you any more to read ! Hope that was of some use.

Answer 3

Let angle A = a,
In tr. ABR & tr. AQC
AB = AQ (sides of equ. tr. ABQ)
AR = AC (sides of equ. tr. ARC)
angle BAR = angle AQC = angle a + 60 degree.
angles of equilateral triangles. BAQ and ARC = 60 degree.
hence two triangles are congruent.
therefore BR = QC ---------------- 1
Similarly in tr. ABP congruent tr. ABR
ttherefore AP = BR --------------2
From 1 and 2 AP = BR = QC.