Gadfly

The maximum possible value for N is 405, in which case AC would have to be 45.

The way I reasoned went like this:

When ABC is divided by AC the quotient is 9 and the remainder 0, so

ABC = 9(AC).

This makes ABC divisible evenly by both 9 and AC. Also, expanding the numbers we get

C+10B+100A =9(C+10A)=9C+90A.

Simplifying, we have,

10(A+B)= 8C, or 5(A+B)= 4C.

Now this last equation implies that A+B is a multiple of 4.

Moreover, A+B cannot be a bigger multiple than 4, since A+B=8 would result in C=10, which it cannot be.

Now since we are looking for the largest N, I set A=4 and B=0, forcing C=5, and low and behold it worked. QED.

Since x > y > z   (meaningx is the largest , y is the second largest, and z is the smallest of the three quantities), then

max(x, y) = x          (since x is the largest),

max(x, z) = x          (since x is the largest),

max(y, z) = y           (since y is the second largest) , and

max(x, y, z) = x      (since x is the largest).

so x + y +z - max(x.y) - max(x, z) - max(y, z) +max(x, y, z) =  x + y + z - x - x - y + x = 2x - 2x +y - y +z =  z . So z remains because he is the smallest.

We have an isosceles quadrilateral; the angles B and C together make up exactly half of 360. OB and OC are bisectors of the angles B and C, and so two of the angles of the triangle OBC add up to 90. So the third should also be 90.

I am also impressed by the amount of work you do and how you stay pleasantly appreciative of other people's work. I could never keep up with you even in the impossible case of having to devote all my time to the forum. And have a happy birthday.

My pleasure, thank you.

Let

\(\alpha= arctan(x)\), and

\(\beta=arctan(x^2)\). Then,

\(tan(\alpha)=x\), and \(tan(\beta)=x^2\), and

\(tan(arctan(x)+arctan(x^2))=Tan(\alpha+\beta)\)

\(=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha)tan(\beta)} =\frac{x+x^2}{1-x\cdot x^2}=\frac{x+x^2}{1-x^3}=x\);

Cross-multiplying , we get

\(x+x^2=x-x^4\), which implies \(x^2=-x^4\). This equation is only true for \(x=0\) (unless we are in the complex field). So there is only one value satisfying the original equation.

In the first blank spot goes

\(\Big\lbrack \begin{matrix} 9\\ 7 \end{matrix} \Big \rbrack \)

In the second blank spot goes

\(-\frac{1}{11} \bigg \lbrack \begin{matrix} 2 &-1\\ -3 &-4 \end{matrix} \bigg\rbrack\)

In the third blank box place

\(-\frac{1}{11} \bigg\lbrack \begin{matrix} 11\\ -55 \end{matrix} \bigg\rbrack\) 

and into the last empty box drag

\( \bigg\lbrack \begin{matrix} -1\\ 5 \end{matrix} \bigg\rbrack\)

Hope this helps you.

A graph is symmetric about the line y=x, if whenever the coordinates of a point (x, y) on the graph are switched,

the resulting point (y, x) would also be on the graph. This means that the transformed equation would be identical to the equation you start with.

Switch x and y in the equation and solve for y:

\(y=\frac{ax+7}{x-4}\) is transformed into \(x=\frac{ay+7}{y-4}\), which when solved for y, gives

\(x(y-4)=1(ay+7)\)       (cross multiply),

\(xy-4x=ay+7\)              (distribute),

\(xy-ay=4x+7\)              (terms containing y on the same side),

\((x-a)y=4x+7\)              (factor out y)

\(y=\frac{4x+7}{x-a}\)                               (divide by x-a)

This equation should be identical to the first equation; so we must have,

\(\frac{ax+7}{x-4}=\frac{4x+7}{x-a}\).

It is clear that we must have \(a=4\)  for these rational expressions to be identical.

In the diagram above r is the radius of the inscribed half-circle and R is the radius of the quarter-circle. I will show that

\(\frac{r^2}{R^2}=\frac{1}{3}\)

so that the ratio of their areas would be

\(\frac{\frac{\pi\cdot r^2}{2} }{\frac{\pi\cdot R^2}{4}}=\frac{2\cdot r^2}{R^2}= \frac{2}{3}\).

(refer to the image above) we have

\(R^2=r^2+(r\cdot \sqrt{2})^2=r^2+2\cdot r^2=3r^2\)

so that

\(\frac{r^2}{R^2}=\frac{1}{3}\) as promised. Q.E.D

Let's say the certain polynomial is \(f(x)\). Then

\(f(x)=g_1(x)\cdot(x-2)+2\), and

\(f(x)=g_2(x)\cdot(x+2)-2\).

so

\(f(2)=g_1(2)\cdot(2-2)+2=2\), and

\(f(-2)=g_2(-2)\cdot(-2+2)-2=-2\)

If we divide \(f(x)\) by \(x^2-4\), we end up with a quotient polynomial \(g(x)\) and remainder \(r(x)\),

where \(r(x)\) is a linear function of x, since it's degree is1 less that the degree of the divisor polynomial. So

we can write

\(f(x)=g(x)\cdot(x^2-4)+r(x)\), where \(r(x)=mx+b\).

Now evaluate the function \(f\) at 2 and -2:

\(f(2)=g(2)\cdot(2^2-4)+r(2)=0+r(2)=2\), which implies

\(r(2)=2m+b=2\), and

\(f(-2)=g(-2)\cdot((-2)^2-4)+r(-2)=r(-2) \), which means

\(r(-2)=-2m+b=-2\).

By eliminating b from the above two equations we get \(m=1 \) and \(b=0\); that gives us the remainder

\(r(x)=x\)

\(OB^2=r^2+x^2\)

\(OC^2=r^2+y^2\)

\(BC^2=OB^2+OC^2=2r^2+x^2+y^2\), and on the other hand

\(BC^2=EC^2+BE^2=(2r)^2+(y-x)^2\). Eliminating \(BC^2\) from the last two equations results in,

\((2r)^2+(y-x)^2=2r^2+x^2+y^2\); simplifying we get

\(r^2=xy\), which implies \(r=\sqrt{xy}\). QED

The equilateral triangles seem to be constructed as in the diagram above:

they are constructed on the three sides of a 30-60-90 right triangle. The leftmost green one on the hypothenuse, the leftmost orange one on the longer leg, and the next green one in the sequence on the shorter leg. So the ratio of the sides of the triangles in the green-orange sequence, I think, is as follows:

The sides are in the ratio \(2:\sqrt{3}:1:\frac{\sqrt{3}}{2} :\frac{1}{2}:\frac{\sqrt{3}}{4}\). So the ratio of the area of the leftmost green to that of the  rightmost

orange should be:

\(\frac{4}{\frac{3}{16}}=\frac{64}{3}\)

If we show that the measure of angle f in the triangle BCF is 90 degrees, then the law of cosines

would tell us that

\(6^2=BF^2+CF^2-2(BF)(CF)cos(f)=BF^2+CF^2-2(BF)(CF)cos (90) \)

\(=BF^2+CF^2\)

since cos(90) = 0.

But f is indeed 90 degrees:

The three interior angles of BFC add up to 180, so

\(f+\alpha +c+\beta+b=180\)

The three interior angles of DBC add up to 180, so

\(\alpha+2b+\beta=80\)

and from the interior angles of EBC we get

\(\beta+2c+\alpha=100\)

Adding the second and third equation we get

\(2\alpha+2\beta+2b+2c=180\), giving us

\(\alpha+\beta+b+c=90\).

This fact added to the first equation results in

\(f=\alpha+\beta+b+c=90\)