practice problems

Q. Point A, inside an acute angle, is reflected in either side of the angle to obtain points B and C. Line segment BC intersects the sides of the angle at D and E (see the Figure). Show that BC/2>DE.

Solution:
The line containing D is perpendicular to AB and it bisects AB as well so it's the perpendicular bisector.
=> BD = AD.
Similarly, AE = EC
BC = BD + DE + EC = AD + DE + AE
In Triangle ADE, AD + AE > DE => AD + AE + DE > 2DE => BC > 2DE H.P.

Q. Two villages lie on opposite sides of a river whose banks are parallel lines. A bridge is to be built over the river, perpendicular to the banks. Where should the bridge be built so that the path from one village to the other is as short as possible?
Solution:
Let the villages be A and B.
Width of the river = W.
Bridge is PQ and P is on A's side and Q is on B's side.
Total distance = AP + PQ + QB
PQ is fixed = W.
So we need to minimize AP + QB.

Hypothetically, shift A to A' s.t. AA' = W(river's width) and AA' is perpendicular to the bank.
Now AA'QP is a ||gram.
And A'Q = AP
So minimizing AP + QB is same as minimizing A'Q + QB.
And best way to minimize A'Q + QB is to make A'QB a straight line.
So if we join a line joining A' and B and take its intersection point with the bank on the side of B as Q then that should be one end of the bridge.

Q. Real numbers a and b are chosen with 1 <a < b such that no triangle with positive area has side lengths 1,a,b and 1,1/a,1/b . What is the smallest possible value of b?

Solution:

Since 1 < a < b and triangle is not possible
=> 1 + a <= b
=> b >= a + 1

Similarly
1 > 1/a > 1/b
=> 1/a + 1/b <= 1
=> 1/b <= 1 - 1/a = (a-1)/a
=> b >= a/(a-1)

For b to satisfy both the inequalities and to be minimum, we need to find intersection of these 2 inequalities.
We can safely equate the inequalities because one is increasing and the other is decreasing.
If both were increasing or decreasing, we will have to look for other ways.
So the minimum b which is >= both will be found at their intersection.
If we check the graphs:

And solve by equating:
a + 1 = a/(a-1)
=> a^2 - 1 = a => a^2 - a - 1 = 0 => a = [1 +- sqrt(5)]/2
Since a > 0 => a = [1 + sqrt(5)]/2
=> b = [3 + sqrt(5)]/2

Q. Point C lies inside a given right angle, and points A and B lie on its sides (see Figure 42). Prove that the perimeter of triangle ABC is not less than twice the distance OC, where O is the vertex of the given right angle.

Working Solution:
Hint: If you think of X and Y axes as mirrors then you can think of light starting from C bouncing off A then to B and then back to C.
Mirrors => reflection
Since C is the only point away from mirrors, let's reflect that.

Pr. th. 
AB + BC + CA >= 2OC
Let's reflect C across OA to C1 and across OB to C2.
OA is perpendicular bisector of CC1 and OB is perpendicular bisector of CC2.
So OC = OC1 = OC2
AC = AC1 and BC = BC2
So pr. th.
AB + BC2 + AC1 >= 2OC
If C is (x,y) then C1 = (-x,y) and C2 = (x,-y)
Their midpoint is (0,0) which is O.
=> C1OC2 is a straight line.
Shortest path between C1 and C2 is of length C1O + OC2 = 2OC
And this will be smaller than any other path for e.g.C1A + AB + BC2. H.P.

Solution that didn't quite work:
Pr. th. AB + BC + AC > 2OC
In triangle AOC,
AO +AC > OC

In triangle BOC,
BO + BC > OC

Add them:
AO + BO + AC + BC > 2OC
In triangle AOB, AO + BO > AB
=> AO + BO + AC + BC > AB + BC + AC

But we have reached a dead end here.

Q. If point O is inside triangle ABC, then prove that AB + BC > AO + OC
Hint:
If you do it using only those triangles which are made up of A,B,C,O then you will eventually hit a dead end. So let's try something else.
We need to show that the outer path from A to C (A-B-C) is bigger than the inner path (A-O-C).
Let's create a middle path to bridge them.
Create a middle path using AOD where D lies on BC.

Solution:
Extend AO to meet BC at D.
Consider triangle ABD.
AB + BD > AD = AO + OD

In triangle ODC:
OD + CD > OC

Add both:
AB + BD + OD + CD > AO + OD + OC
Cancel OD
AB + BD + DC > AO + OC
=> 
AB + BC > AO + OC
H.P.

Q. Prove that the length of median AM in triangle ABC is not greater than half the sum of sides AB and AC. Prove also that the sum of the lengths of the three medians is not greater than the triangle's perimeter.

Hint:
AM <= (AB + AC)/2
=> 2AM <= (AB + AC)
2AM gives a hint that we need to extend AM to make it double.
Also we need a triangle where AB,AC,2AM co-exist. Which can be done by translating AC to a parallel side next to AB.
Solution:
Extend the median AM to AD s.t. AM = MD
ABDC is a ||gram. Why? Its diagonals BC and AD bisect each other.
So AB = CD and AC = BD
In triangle ABD, AB + BD > AD, similarly AC + CD > AD
Add both and substitute to get AB + AC > 2AM.
Do it for all the medians to get 2*perimeter > 2*(sum of medians)
H.P.

Q. A fly sits on one vertex of a wooden cube. What is the shortest path it can follow to the opposite vertex?
Answer: sqrt(5) units.
Solution:
First solution we think of is 3 units. One edge in each dimension.
Second solution is to traverse diagonal of the current face and then move 1 unit: sqrt(2) + 1 units.
Even better is to mentally unfold 2 faces and make them one rectangle of 2x1 and traverse its diagonal: sqrt(5) units.