Find the infinite product

## Monday Math 161

September 15, 2014## Monday Math 160

September 8, 2014Suppose we have four identical-looking coins. Three are fair, but one is biased, with a probability of coming up heads of 3/5. We select one of the four coins at random.

1. If we flip the selected coin twice, and it comes up heads both times, what is the probability that our coin is the biased one?

2. If we flip the selected coin three times, and it comes up heads all three times, what, then, is the probability that our coin is the biased one?

3. Generalize: We have *m* fair coins and one identical-looking biased coin with probability *p* of getting heads. If we select one coin at random, and obtain *k* heads in *n* flips, what is the probablility P(*m*,*p*,*n*,*k*) that we have the biased coin?

## Monday Math 159

August 18, 2014Find

## Monday Math 158

August 11, 2014Find a non-summation expression for the value of the sum .

## Monday Math 157

August 4, 2014Consider a triangle, which we label ∆*ABC*, with circumcenter *O* and circumradius *R*=*AO*=*BO*=*CO*. Let us label the midpoints of the sides as *M _{A}*,

*M*and

_{B}*M*, so that

_{C}*M*is the midpoint of

_{A}*BC*(the side opposite

*A*), and similarly, so that , and are the medians. Then , < and are segments of the perpendicular bisectors of the sides. Let us also label the feet of the altitudes from

*A*,

*B*and

*C*as

*H*,

_{A}*H*and

_{B}*H*, respectively, and let

_{C}*H*be the orthocenter (the intersection of the altitudes , and ).

Let us construct the point *D* on the circumcircle diametrically opposed to *A*; that is to say, the point *D* such that *AD* is a diameter of the circumcenter. Then *AD*=2*R*, and *O* is the midpoint of *AD*.

Now, by Thales’ theorem, ∠*ABD* and ∠*ACD* are both right angles. Now, since *CD* and the altitude are both perpendicular to *AC*, they are parallel to each other. Similarly the altitude and segment *BD* are parallel, both being perpendicular to *AB*. Thus, the quadrilateral *BDCH* is a parallelogram.

Since the diagonals of a parallelogram bisect each other, we see that the midpoint *M _{A}* of

*BC*is also the midpoint of

*HD*.

Now, let *P _{A}* be the midpoint of the segment

*AH*. Then is a midline of the triangle ∆

*AHD*, and by the triangle midline theorem, and

*P*=½

_{A}M_{A}*AD*=

*R*.

Now, let *N* be the intersection of and *HO*. By the midline-median bisection theorem proven in this post, we see that, as *HO* is the median of ∆*AHD* that crosses midline , *N* is the midpoint of both and *HO*. Thus, *NM _{A}*=

*NP*=½

_{A}*P*=½

_{A}M_{A}*R*.

Now, consider the quadrilateral *HOM _{A}H_{A}*. Since and are both perpendicular to

*BC*,

*HOM*is a right trapezoid. Letting

_{A}H_{A}*Q*be the midpoint of , we see then that

_{A}*NQ*is the median (or midline) of trapezoid

*HOM*.

_{A}H_{A}By the first of the three items proven here, we see that , and so . Thus, we see that is the perpendicular bisector of , and so, by the perpendicular bisector theorem, *NH _{A}*=

*NM*, and so

_{A}*NH*=

_{A}*NM*=

_{A}*NP*=½

_{A}*R*.

Constructing diameter *BE* of the circumcircle gives us parallelogram *CEAH*, by a similar argument as above. Letting *P _{B}* be the midpoint of the segment

*BH*, analogous reasoning to the above shows that

*NH*=

_{B}*NM*=

_{B}*NP*=½

_{B}*R*as well. Lastly, diameter

*CF*, and midpoint

*P*of

_{C}*CH*gives, by similar proof, that

*NH*=

_{C}*NM*=

_{C}*NP*=½

_{C}*R*. Thus, the nine points

*M*,

_{A}*M*,

_{B}*M*,

_{C}*H*,

_{A}*H*,

_{B}*H*,

_{C}*P*,

_{A}*P*and

_{B}*M*are all equidistant from

_{C}*N*.

Therefore, we see that for any triangle, the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments connecting the vertices to the orthocenter all lie on a single circle, which, for this reason, is usually known as the nine-point circle, and its center *N* the nine-point center. We also see that the radius of the the nine-point circle is half the radius of the circumcircle, and the nine-point center is the midpoint of the segment connecting the circumcenter and orthocenter. This latter tells us that for any non-equilateral triangle, the nine-point center lies on the Euler line (for an equilateral triangle, it coincides with the circumcenter, orthocenter, and centroid).

## Monday Math 156

July 28, 2014For a trapezoid, the median (also known as the midline or mid-segment) is the segment connecting the midpoints of the legs of the trapezoid. This post presents a proof that:

- the median of a trapezoid is parallel to the bases;
- the length of the median is half the sum of the lengths of the bases;
- the midpoints of the diagonals of a trapezoid also lie on its midline.

## Monday Math 155

July 21, 2014The triangle midline theorem, also called the midsegment theorem, states that the line segment connecting the midpoints of two sides of a triangle is parallel to and half the length of the third side. This may be trivially proven via triangle similarity (SAS similarity condition) and the corresponding angles postulate. More interesting, however, is to prove it using triangle congruence.

Let *D* and *E* be the midpoints of sides *AB* and *AC*, respectively, of ∆*ABC*. Let us extend segment *DE* past *E* to point *F* such that *DE*=*EF*, and let us draw *CF*.

Since *DE*=*EF*, *AE*=*EC*, and vertical angles ∠*AED* and ∠*CEF* are congruent, we see by the SAS condition that ∆*ADE*≅∆*CFE*. Thus, *CF*=*AD*=*BD*. Also, ∠*FCE*≅∠*DAE*; but since these are alternate interior angles for lines and cut by transversal , we see that . But then the quadilateral *BCFD* has a pair of opposite sides, *BD* and *CF*, which are of equal length and parallel, so it is therefore a parallelogram, and so . And since opposite sides of a parallelogram have equal length, *DF*=*BC*, and so *DE*=½*DF*=½*BC*.

We can also prove a similar theorem: that the line through the midpoint of one side of a triangle parallel to a second side of the triangle bisects the third side, and that the segment of that line inside the triangle is one-half as long as the parallel side. As with the midpoint theorem, it is trivial to prove with triangle similarity (this time the AA condition) and the corresponding angles postulate. So, instead, we use a similar construction as above.

Let *D* be the midpoint of side *AB* of ∆*ABC*, and let *E* be the point where the line through *D* parallel to *BC* intersects *AC*. Construct the line through *C* parallel to *AB*, and let *F* be the point where it intersects

Then *BCFD* is a parallelogram, and since opposite sides of a parallelogram have the same length, *BD*=*CF* and *BC*=*DF*. And so *CF*=*BD*=*AD*. And since ∠*ADE* and ∠*CFE* are alternate interior angles for lines and cut by transversal , they are congruent. Similarly, ∠*DAE*≅∠*CFE*, and so, by the ASA condition, ∆*ADE*≅∆*CFE*. Thus, *AE*=*EC*, and *E* is thus the midpoint of *AC*. We see also that *DE*=*EF*, and since *DF*=*BC*, thus *DE*=½*DF*=½*BC*.

Using both of these theorems together, we can prove a third: that a midline (mid-segment) of a triangle and the triangle median that intersects it bisect each other.

Let *M _{A}*,

*M*and

_{B}*M*be the midpoints of sides

_{C}*BC*,

*AC*and

*AB*, respectively, of ∆

*ABC*. Thus, is a midline of ∆

*ABC*, and a median. Let

*P*be the point where they intersect.

By the midline theorem, and *M _{B}M_{C}*=½

*BC*.

This means, then, that , and so by our second theorem above with regards to ∆*ABM _{A}*, we see that

*P*must be the midpoint of , and

*M*=½

_{C}P*BM*.

_{A}Similarly, our second theorem applied to triangle *AM _{A}C* establishes that

*M*=½

_{B}P*CM*. But

_{A}*BM*=

_{A}*CM*, and so

_{A}*M*=

_{B}P*M*, and

_{C}P*P*is the midpoint of as well.

## Monday Math 154

July 14, 2014Continuing the series on triangle centers, let us consider ∆*ABC*, with circumcenter *O* and centroid *G*. Construct the line segment *OG*, and extend it out from *G* to the point *H* such that *GH*=2*OG*.

Next, construct the median from vertex *A* to the midpoint *M* of side *BC*. Then *G* lies on *AM*, with *AG*=2*GM*, as we showed here. Recalling that the circumcenter is the intersection of the perpendicular bisectors of the sides, we see that *OM*⊥*BC*.

Now, since *AG*=2*GM*, *GH*=2*OG*, and ∠*AGH*≅∠*MGO*, we see (by the SAS similarity condition) that ∆*AGH*~∆*MGO*. And since these triangles are similar, corresponding angles ∠*HAG* and ∠*OMG* are congruent. However, these are alternate interior angles for lines and cut by transversal , and therefore . And since *OM*⊥*BC*, we see , and is the triangle altitude from *A* to *BC*.

Analogous constructions show that *H* must also be on the triangle altitudes from *B* and *C*:

Thus, we see that *H* is the orthocenter of ∆*ABC*. So, we see that for any non-equilateral triangle, the circumcenter *O*, centroid *G* and orthocenter *H* are collinear, with *GH*=2*OG*; the line through these triangle centers is known as the Euler line of the triangle. (For an equilateral triangle, *O*, *G* and *H* are all the same point.)

## Monday Math 153

July 7, 2014A few weeks ago, I demonstrated a proof that all three medians of any triangle are concurrent; in other words, that the centroid exists.

Now, how about demonstrating that the circumcenter and orthocenter exist; that is to say, showing that the three perpendicular bisectors of the sides are concurrent, and that the three altitudes are concurrent, for any triangle.