practice problems

Q1. Let X be the midpoint of the side AB of △ABC. Let Y be the midpoint of CX. Let BY cut AC at Z. Prove that AZ=2ZC.

Q2. ABC is a equilateral triangle with vertex A fixed and B moving in a given straight line. Find the locus of C.

Q3.

Let A be one of the two points of intersection of two circles with centres X and Y. The tangents at A to these two circles meet the circles again at B,C. Let the point P be located so that PXAY is a parallelogram. Show that P is the circumcentre of triangle ABC.

Q4:
Let ABC be a triangle and h be the altitude from A to BC.
Prove: (b+c)^2 >= a^2 + 4h^2

Q5:
The interior of a quadrilateral is bounded by the graphs of $(x+ay{)}^2=4a^2$ and $(ax-y{)}^2=a^2$, where $a$ is a positive real number. What is the area of this region in terms of $a$, valid for all $a>0$ ?

Q6. Three equally spaced parallel lines intersect a circle, creating three chords of lengths 38, 38, and 34. What is the distance between two adjacent parallel lines?


Solution 1:
Solution A:
Mass point geometry.
Let weight of C Be 2, since Y is m.p. of CX, weight of X is also 2.
X is m.p. of AB so weight of A,B is 1 each.
Now weight of A is 1, C is 2 so AZ = 2CZ, H.P.

Solution B:
Draw a line from X parallel to BZ. Let it cut AC at W.

In triangle ABZ, W is midpoint of AZ because X is mp of AB and XW || BZ.
IN triangle CXW, YZ || XW => Z is m.p. of CW. H.P.

Solution C:
Menelaus's theorem:
BYZ is a transversal for triangle AXC, and B sits on AX extended.
AB/BX * XY/YC * CZ/ZA = 1 = 2 * 1 * CZ/ZA => ZA = 2CZ H.P.

Solution 2:
B moves in a fixed line and AC is nothing but AB rotated by 60 degree clockwise or counterclockwise.
So locus of C will be 2 lines which make 60 degree angle with B's locus line in 2 directions.

Solution 3:
To show that P is the Circumcenter we can either show that it's equidistant from all 3 vertices of ABC or that it lies on 2 of the perpendicular bisectors. We will go with the second option.

YA is perpendicular to AB and XA is perpendicular to AC.
Since PXAY is a ||gram YA || XP and XP is also perpendicular to AB.
X is the center and AB is a chord.
Perpendicular from center to chord bisects it.
=> XP is the perpendicular bisector of AB.

Similarly PY is the perpendicular bisector of AC.
And hence P is the circumcenter of ABC.

Solution 4:
Since there are squares of a,b,c we can try cosine rule.
b^2 + c^2 - a^2 = 2bc.cosA
Add 2bc.
(b+c)^2 - a^2 = 2bc(1+cosA)

Area of triangle = (bc.sinA)/2 = (h.a)/2 => h = (bc.sinA)/a => 4h^2 = 4.b^2.c^2.sin^2A/a^2
To prove that
(b+c)^2 - a^2  = 2bc(1+cosA) >=  4.b^2.c^2.sin^2A/a^2
=> (1 + cosA) >= 2bc.(1-cosA)(1+cosA)/a^2
=> 1 >= 2bc(1-cosA)/a^2 (we can cancel 1+cosA since cosA is (-1,1) for a triangle angle and hence positive.)
=> a^2 >= 2bc(1-cosA)
=> b^2 + c^2 - 2bc.cosA >= 2bc(1-cosA)
=> b^2 + c^2 >= 2bc
=> (b-c)^2 >= 0 H.P.

Solution 5:
ax - y = +-a
and
x + ay = +-2a
are the parallel lines forming this quadrilateral.
And they are also perpendicular.
So it's a rectangle.
So area = width*height

Distance between 2 parallel lines = 
|C1 - C2|/sqrt(A^2 + B^2)
where the lines are
Ax + By = C1
Ax + By = C2

So width = 
2a/sqrt(1+a^2)

height = 
4a/sqrt(1 + a^2)

Area = 8a^2/(1 + a^2)

Solution 6:
Chords having length 38 are equidistant from center and the 34 one is below one of them.
Let the center's distance from 38 chord be x.
R^2 - x^2 = 19^2
R^2 - (3x)^2 = 17^2
8x^2 = 19^2 - 17^2
=> x = 3 => 2x = 6 = answer.