answers

problem 10

2011-11-29T05:05:00.001-08:00

the question was
Let be a finite set of at least two points in the plane. Assume that no three points of are collinear. A windmill is a process that starts with a line going through a single point . The line rotates clockwise about the pivot until the first time that the line meets some other point belonging to . This point, , takes over as the new pivot, and the line now rotates clockwise about , until it next meets a point of . This process continues indefinitely.
Show that we can choose a point in and a line going through such that the resulting windmill uses each point of as a pivot infinitely many times.

now lets see its solution

This problem was very nice, creative and inspiring. Something a bit new which looks very nice. I will present a solution I found on AoPS, which I consider very interesting.

For any point , there is a line which cuts the remaining points in two groups, which differ with at most point from one another. This kind of a line works for the problem. Why? Imagine how such a line creates the windmill: it touches one point, then one side gains one point, and then this side loses the pivot, leaving the situation as before, i.e. with sides of the line differing by at most one point.

Pick now one point and a line which has the above property. At a given moment, the windmill will be parallel to this line, and by uniqueness, will coincide with this line. This reasoning shows that any point will be pivot in a rotation of the windmill.

question 8

2011-11-24T05:04:00.001-08:00

Problem was this:

In the interior of triangle a point is given. Let be the intersections of with the opposing edges of triangle . Prove that among the ratios there exists one not larger than and one not smaller than .

Solution

Since triangles and share the base , we have , where denotes the area of triangle . Similarly, . Adding all of these gives , or We see that we must have at least one of the three fractions less than or equal to , and at least one greater than . These correspond to ratios being less than or equal to , and greater than or equal to , respectively, so we are done.

question 7

2011-11-23T05:57:00.001-08:00

question was

Find all function such that for all the following equality holds

where is greatest integer not greater than

Solutions

Solution 1

Put . Then or .

If , putting we get , that is f is constant. Substituing in the original equation we find or , where .

If , putting we get or .

For , we set to find , which is a solution.

For , setting yields .

Putting to the original we get . However, from we have , so which contradicts the fact .

So, or .

Solution 2

Substituting we have . If then . Then is constant. Let . Then substituting that in (1) we have , or . Therefore where or

If then . Now substituting we have . If then and substituting this in (1) we have . Then . Substituting we get . Then , which is a contradiction Therefore . and then for all

Then the only solutions are or where .( By m.candales [2])

Solution 3

Let , then .

Case 1:

Then is a constant. Let , then . It is easy to check that this are solutions.

Case 2:

In this case we conclude that

Lemma:If is such that ,

Proof of the Lemma: If we have that , as desired.

Let , so that we have: , using the lemma.

If is not constant and equal to , letting be such that implies that .

Now it's enough to notice that any real number is equal to , where and , so that . Since was arbitrary, we have that is constant and equal to .

We conclude that the solutions are , where .( By Jorge Miranda

Solution 4

Clearly , so for all .

If for all , then by taking we get , so is identically null (which checks).

If, contrariwise, for some , it follows for all .

Now it immediately follows , hence .

For this implies . Assume ; then , absurd.

Therefore , and now in the given functional equation yields for all , therefore constant, with , i.e. (which obviously checks).

question 6

2011-11-22T00:06:00.001-08:00

the question was

Let  be a point interior to triangle  (with ). The lines ,  and  meet again its circumcircle  at , , respectively . The tangent line at  to  meets the line  at . Show that from  follows .

solution

Without loss of generality, suppose that . By Power of a Point, , so  is tangent to the circumcircle of . Thus, . It follows that after some angle-chasing,
so  as desired.

Solution 2

Let the tangent at to intersect at . We now have that since and are both isosceles, . This yields that .

Now consider the power of point with respect to .

Hence by AA similarity, we have that . Combining this with the arc angle theorem yields that . Hence .

This implies that the tangent at is parallel to and therefore that is the midpoint of arc . Hence .

question 5

2011-11-14T21:13:00.001-08:00

the question was :

Given any set of four distinct positive integers, we denote the sum by . Let denote the number of pairs with for which divides . Find all sets of four distinct positive integers which achieve the largest possible value of .

answer
Solution:
Firstly, if we order , we see , so isn't a couple that satisfies the conditions of the problem. Also, , so again isn't a good couple. We have in total 6 couples. So .
We now find all sets with . If and are both good couples, and , we have . So WLOG with and . It's easy to see and since are bad, all couples containing must be good. Obviously and are good (). So we have and .
Using the second equation, we see that if , , for some a positive integer.
So now we use the first equation to get , for a natural .
Finally, we obtain 1, 2 or 4. We divide in cases:
CASE I: . So and . But 3, 4,5 or 6. implies , impossible. when . We easily see and , impossible since . When , , and we get .Uf , and we get .
CASE II and III:2, 4. Left to the reader.
ANSWER: ,, for any positive integer .

it was a very hard question seems to be if you solve you are brilliant

question 4

2011-11-14T06:48:00.001-08:00

Question was :-
Let ABCDEF be a convex hexagon in which the diagonals AD, BE, CF are concurrent
at O. Suppose the area of traingle OAF is the geometric mean of those of OAB and
OEF; and the area of triangle OBC is the geometric mean of those of OAB and OCD.
Prove that the area of triangle OED is the geometric mean of those of OCD and OEF.

Solution:
Let OA = a
OB = b,
OC = c,
OD = d,
OE = e,
OF = f,
[OAB] =x,
[OCD] = y,
[OEF] = z,
[ODE] = u,
[OFA] = v
[OBC]= w.
We are given that v^2 = zx, w^2 = xy and we have to prove that u^2 =yz.
Since∠AOB =∠DOE, we have
u\x =(1/2 de sin∠DOE)/(1/2 ab sin∠AOB)= de/ab

Similarly, we obtain
v/y =fa/cd
w/z =bc/ef
Multiplying, these three equalities, we get uvw =xyz. Hence
x^2 * y^2 * z^2 =u^2*v^2*w^2 =u^2 (zx)(xy).
This gives u^2=yz, as desired.

question 3

2011-11-13T23:32:00.001-08:00

Question was
Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?

Explanatory Answer
There are three possible cases that will satisfy the condition of forming three letter passwords with at least 1 symmetric letter.

Case 1: 1 symmetric and 2 asymmetric or
Case 2: 2 symmetric and 1 asymmetric or
Case 3: all 3 symmetric

= {(¹¹C₁ * ¹⁵C₂) + (¹¹C₂ * ¹⁵C₁) + ¹¹C₃} * 3!

= {11 * + * 15 + } x 6

= {1155 + 825 + 165} * 6

= 2145 * 6 = 12870

i hope you liked that

Question 2

2011-11-13T22:00:00.001-08:00

the question was

Let ABC be a triangle in which AB = AC and let I be its in-centre. Suppose
BC = AB+AI. Find∠BAC

Solution:

We observe that∠AIB = 90◦ +(C/2). Extend CA to D such that AD = AI.
Then CD =CB by the hypothesis. Hence∠CDB =∠CBD = 90◦−(C/2). Thus
∠AIB+∠ADB = 90◦+(C/2)+90◦−(C/2)= 180◦.
Hence ADBI is a cyclic quadrilateral. This implies that
∠ADI =∠ABI = B
2.
But ADI is isosceles, since AD = AI. This gives
∠DAI = 180◦−2(∠ADI)= 180◦−B.
Thus∠CAI = B and this gives A = 2B. Since C = B, we obtain 4B = 180◦ and
hence B = 45◦. We thus get A= 2B = 90◦.

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question 1

2011-11-13T21:45:00.001-08:00

the question was

a^2 + 2b = 7
b^2 + 4c = -7
c^2 + 6a = -14
then a^2 + b^2 + c^2 = ?
(a)14
(b)25
(c)36
(d)47

the solution is

to solve add all equations getting u to:
a^2+6a+b^2+2b+c^2+4c= -14
=> (a^2+6a+9)+(b^2+2b+1)+(c^2+4c+4) = 0
=>(a+3)^2+(b+1)^2+(c+2)^2 = 0
=> a=-3 , b=-1 , c=-2
=>a^2+b^2+c^2=14

I hope you have understand the question

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