*Determine the unique positive two-digit integers *m*
and *n* for which the approximation is accurate to seven decimals, i.e., .*

* *

Let us first take the reciprocal of our approximation for ,
that is, . We get, to the nearest 7^{th} digit,
4.2941179.

So, .

Again, we take the reciprocal of 0.2941179. This is approximately 3.3999974.

Now, .

Repeating these steps, we get

.

Simplifying this continued fraction, and neglecting the last number, we get

.

We verify that , which is indeed accurate to seven decimals, as required.

We also note that the first few convergents for 0.2328767 are

…

The next one has a numerator greater than 100, and is accurate only to four decimal places. Thus, is the unique answer.

*It is well known that there are infinitely many
triples of integers (*a, b, c*) whose greatest common divisor is 1 and
which satisfy the equation**. Prove that there are also infinitely many
triples of integers (*r, s, t*) whose greatest common divisor is 1 and
which satisfy the equation **.*

* *

Let us assume that *r, s, *and *t* are all greater
than 0 (Not necessary integers), and if:

then dividing both sides by *r ^{2}s^{2}t^{2}*
we get

.

Now if this has a solution where *r, s, *and* t*
are rational numbers (not necessarily
integers) we can easily find an
equivalent integer solution. (We multiply each term by a suitable number, as
explained below)

But we do have such a solution, where ,
,
and , and *a*,
*b* and *c* are integer solution of
!

In other words, suppose we have a primitive Pythagorean
triplet (*a, b, c*).

Let ,

Now T, R, and S are
not integers, but they are solutions of *R ^{2}S^{2} *+

Then *kR, kS*,
and *kT* are also solutions because if we multiply the above equation by k^{4}
we get *(kR) ^{2}(kS)^{2} *+

Therefore, to find integer solutions, we
choose: *k* = *abc*,

So that makes *kR,
kS*, and *kT* each an integer.

**So, { r = ac, s = ab, t = bc}
are the unique integer solutions for .**

Also, we will prove that if (*a, b, c*) has a greatest
common divisor of 1, then so does

(*r, s, t*).

If (*a, b*) have a common
divisor *k* > 1, then let *a* = *kx* and *b *=* ky*.

Now, * k ^{2}x^{2}
+ k^{2}y^{2} *=

This implies that *k*
is also a factor of *c*, which means that (*a, b, c*) is not a
primitive triple.

Similarly, any common factor of (*a, c*) will also be a
factor of *b*.

Thus, no pair of (*a, b, c*) has any common factors
greater than 1.

So, GCF(*r, s*) = *a*,

But there are no common factors greater than 1 with *a*
and *bc*

Thus, GCF(*r, s, t*) = 1.

Since there are infinitely many solutions for (*a, b, c*),
there are also infinitely many solutions for (*r, s, t*) by selecting *r
*=* ac, s *=* ab, *and* t *=* bc.*

For example:

(3,4,5) gives a solution of (15, 12, 20), and

(5, 12, 13) gives us (65, 60, 156)

* *

*Suppose ** for some angle *x*, . Determine ** for the same *x*.*

* *

It is easy to see that *x* ≠ 0, or π/2, so
neither sin *x* or cos *x* is 0, hence it permissible to divide by
sin *x* or cos *x* in all the following equations.

We will use

and

=

.

or, (eq 1)

Similarly,

(eq 2)

So, using eq 1 and 2, and the fact that , we get

Since we are given that

*The *projective plane* of order three consists of
13 “points” and 13 “lines”. These lines
are not Euclidean straight lines; instead they are sets of four points with the
properties that each pair of lines has exactly one point in common, and each
pair of points has exactly one line that contains both points. Suppose that points are labeled 1 through 13,
and six of the lines are: A = {1, 2, 4, 8}, B = {1, 3, 5, 9}, C = {2, 3, 6,
10}, D = {4, 5, 10, 11}, E = {4, 6, 9, 12}, and F = {5, 6, 8, 13}. What is the line that contains 7 and 8?*

* *

We know that set *G* contains
7 and 8. Since each pair of points has
only one line which contains both points, we know that no point which occurred
in any other set that contained 7 or 8 may be contained in *G*.

Set *A* contains {1, 2, 4,
8}. Set *F* contains {5, 6, 8,
13}. This means that Set *G* must
not have 1, 2, 4, 5, 6, or 13 as members.
Thus, the only possible choices for the other two members are 3, 9, 10,
11.

The other two members can be 3, 9,
10, 11, and 12. One of these two points
must be common to sets *B, C, D, *and* E*. However, these points must not *both*
occur in any of these sets.

Checking all the possible combinations of these numbers: (3, 9), (3, 10), (3, 11),

(3, 12), (9, 10), (9, 11), (9, 12), (10, 11), (10, 12), and (11, 12) for this, we find that the possible pairs of points that do not both occur in any of the forementioned sets are:

(3, 11), (3, 12,) (9, 10), (9, 11), and (10, 12).

(3, 11) is not a possibility
because set *E* contains neither of them.

(3, 12) is not a possibility
because set *D* contains neither of them.

(9. 10) is a possibility because one of them
is contained in each of the sets *B, C, D, E, *and *F*.

(9, 11) is not a possibility
because neither of them is contained in set *C*.

(10, 12) is not a possibility
because neither of them is contained in set *B*.

The only possible pair of points that can be in a set with 7 and 8 is (9, 10)

**Thus, the only possible set G
is {7, 8, 9, 10}**

* *

* *

* *

*In PQR,
QR < PR < PQ *

*so that the exterior angle **x*
V

*bisector through P intersects *

** ray at point S, and the ** U

*exterior angle bisector at R *

*intersects ray at *

*point T, as shown on the right. *

*Given that PR = PS = RT, *

*determine, with proof, the *

*measure of PRQ.*

Let point *U* lie on in the opposite direction of *T* from *P*.

Let point V lie on in the opposite direction of *P* from *R*.

Let *m**PRQ*
= *x*.

because it is supplementary to *PRQ*.

Also, because *PR *=* PS*.

From we get

, or

*RPQ*
is supplementary to *RPU*,
so

, or

Since *RP* = *RT*,

So,

*QRV*
is supplementary to *PRQ*,
so *m**QRV*
= 180 *x*.

Thus,

From Δ*QRT*
we get

, or

Now, from Δ*PQR*,
we get

, or

**Thus, mPRQ
= 132º**