III. CAPITULO III: LINEA BASE
3. DIAGNOSTICO AMBIENTAL - LINEA BASE
3.2. DESCRIPCION DE LAS INSTALACIONES Y ACTIVIDADES
Problem 1 In the fraction
29 ÷ 28 ÷ 27 ÷ · · · ÷ 16 15 ÷ 14 ÷ 13 ÷ · · · ÷ 2
parentheses may be repeatedly placed anywhere in the numerator, granted they are also placed on the identical locations in the denom- inator.
(a) Find the least possible integral value of the resulting expression. (b) Find all possible integral values of the resulting expression.
Solution:
(a) The resulting expression can always be written (if we refrain from canceling terms) as a ratio A
B of two integers A and B satisfying
AB = (2)(3) · · · (29) = 29! = 225·313·56·74·112·132·17·19·23·29.
(To find these exponents, we could either count primes directly factor by factor, or use the rule that
n p + n p2 + n p3 + · · · (1) is the exponent of p in n!.)
The primes that have an odd exponent in the factorization of 29! cannot “vanish” from the ratio A
B even after making any
cancellations. For this reason no integer value of the result can be less than
H = 2 · 3 · 17 · 19 · 23 · 29 = 1, 292, 646. On the other hand,
29 ÷ (28 ÷ 27 ÷ · · · ÷ 16) 15 ÷ (14 ÷ 13 ÷ · · · ÷ 2) = 29 · 14 15 · 28· (27)(26) · · · (16) (13)(12) · · · (2) = 29 · 14 2 28 · 27! (15!)2 = 29 · 7 · 2 23· 313· 56· 73· 112· 132· 17 · 19 · 23 (211· 36· 53· 72· 11 · 13)2 = H.
(Again it helps to count exponents in factorials using (1).) The number H is thus the desired least value.
(b) Let’s examine the products A and B more closely. In each of the fourteen pairs of numbers
{29, 15}, {28, 14}, . . . , {16, 2},
one of the numbers is a factor in A and the other is a factor in B. The resulting value V can then be written as a product
29 15 1 28 14 2 · · · 16 2 14 ,
where each iequals ±1, and where 1= 1 and 2= −1 no matter
how the parentheses are placed. Since the fractions 2713, 2612, . . . ,
16
2 are greater than 1, the resulting value V (whether an integer
or not) has to satisfy the estimate V ≤ 29 15· 14 28· 27 13· 26 12· . . . · 16 2 = H,
where H is number determined in part (a). It follows that H is the only possible integer value of V !
Problem 2 In a tetrahedron ABCD we denote by E and F the
midpoints of the medians from the vertices A and D, respectively. (The median from a vertex of a tetrahedron is the segment connecting the vertex and the centroid of the opposite face.) Determine the ratio of the volumes of tetrahedrons BCEF and ABCD.
Solution: Let K and L be the midpoints of the edges BC and AD, and let A0, D0 be the centroids of triangles BCD and ABC,
respectively. Both medians AA0 and DD0 lie in the plane AKD,
and their intersection T (the centroid of the tetrahedron) divides them in 3 : 1 ratios. T is also the midpoint of KL, since ~T =
1 4( ~A + ~B + ~C + ~D) = 1 2( 1 2( ~A + ~D) + 1 2( ~B + ~C)) = 1 2( ~K + ~L). It follows
that ETAT = F TDT = 13, and hence 4AT D ∼ 4ET F and EF = 13AD. Since the plane BCL bisects both segments AD and EF into halves, it also divides both tetrahedrons ABCD and BCEF into two parts of equal volume. Let G be the midpoint of EF ; the corresponding volumes than satisfy
[BCEF ] [ABCD] = [BCGF ] [BCLD] = GF LD· [BCG] [BCL] = 1 3 KG KL = 1 3· 2 3 = 2 9.
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Problem 3 Show that there exists a triangle ABC for which, with the usual labelling of sides and medians, it is true that a 6= b and a + ma = b + mb. Show further that there exists a number k such
that for each such triangle a + ma = b + mb= k(a + b). Finally, find
all possible ratios a : b of the sides of these triangles.
Solution: We know that
m2a = 1 4(2b 2+ 2c2− a2), m2 b = 1 4(2a 2+ 2c2− b2), so m2a− m2 b = 3 4(b 2− a2).
As ma− mb = b − a 6= 0 by hypothesis, it follows that ma+ mb = 3
4(b + a). From the system of equations
ma− mb= b − a
ma+ mb =
3 4(b + a) we find ma= 18(7b − a), mb=18(7a − b), and
a + ma= b + mb=
7
8(a + b). Thus k = 78.
Now we examine for what a 6= b there exists a triangle ABC with sides a, b and medians ma= 18(7b − a), mb =18(7a − b). We can find
all three side lengths in the triangle AB1G, where G is the centroid
of the triangle ABC and B1 is the midpoint of the side AC:
AB1= b 2, AG = 2 3ma = 2 3 · 1 8(7b − a) = 1 12(7b − a), B1G = 1 3mb= 1 3 · 1 8(7a − b) = 1 24(7a − b).
Examining the triangle inequalities for these three lengths, we get the condition
1 3 <
a b < 3,
from which the value ab = 1 has to be excluded by assumption. This condition is also sufficient: once the triangle AB1G has been
constructed, it can always be completed to a triangle ABC with b = AC, ma = AA1, mb = BB1. Then from the equality m2a− m2b =
3 4(b
Problem 4 In a certain language there are only two letters, A and B. The words of this language satisfy the following axioms:
(i) There are no words of length 1, and the only words of length 2 are AB and BB.
(ii) A sequence of letters of length n > 2 is a word if and only if it can be created from some word of length less than n by the following construction: all letters A in the existing word are left unchanged, while each letter B is replaced by some word. (While performing this operation, the B’s do not all have to be replaced by the same word.)
Show that for any n the number of words of length n equals 2n+ 2 · (−1)n
3 .
Solution: Let us call any finite sequence of letters A, B a “string.” From here on, we let · · · denote a (possibly empty) string, while ∗ ∗ ∗ will stand for a string consisting of identical letters. (For example, B ∗ ∗ ∗ B
| {z }
k
is a string of k B’s.)
We show that an arbitrary string is a word if and only if it satisfies the following conditions: (a) the string terminates with the letter B; and (b) it either starts with the letter A, or else starts (or even wholly consists of) an even number of B’s.
It is clear that these conditions are necessary: they are satisfied for both words AB and BB of length 2, and they are likewise satisfied by any new word created by the construction described in (ii) if they are satisfied by the words in which the B’s are replaced.
We now show by induction on n that, conversely, any string of length n satisfying the conditions is a word. This is clearly true for n = 1 and n = 2. If n > 2, then a string of length n satisfying the conditions must have one of the forms
AA · · · B, AB · · · B, B ∗ ∗ ∗ B | {z } 2k A · · · B, B ∗ ∗ ∗ B | {z } 2k+2 ,
where 2 ≤ 2k ≤ n−2. We have to show that these four types of strings arise from the construction in (ii) in which the B’s are replaced by strings (of lengths less than n) satisfying the condition — that is, by words in view of the induction hypothesis.
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The word AA · · · B arises as A(A · · · B) from the word AB. The word AB · · · B arises either as A(B · · · B) from the word AB, or as (AB)(· · · B) from the word BB, depending on whether its initial letter A is followed by an even or an odd number of B’s. The word B ∗ ∗ ∗ B
| {z }
2k
A · · · B arises as (B ∗ ∗ ∗ B)(A · · · B) from the word BB, and the word B ∗ ∗ ∗ B
| {z }
2k+2
as (B ∗ ∗ ∗ B
| {z }
2k
)(BB) from the word BB. This completes the proof by induction.
Now we show that the number pn of words of length n is indeed
given by the formula
pn=
2n+ 2 · (−1)n
3 .
It is clearly true for n = 1 and 2 since p1 = 0 and p2 = 2; and the
formula will then follow by induction if we can show that pn+2 =
2n+ p
n for each n. But this recursion is obvious because each word
of length n + 2 is either of the form A · · · B where · · · is any of 2n strings of length n; or of the form BB · · · where · · · is any of the pn
words of length n.
Problem 5 In the plane an acute angle AP X is given. Show how to construct a square ABCD such that P lies on side BC and P lies on the bisector of angle BAQ where Q is the intersection of ray P X with CD.
Solution: Consider the ration by 90◦ around the point A that maps B to D, and the points P, C, D into some points P0, C0, D0, respectively. Since ∠P AP0 = 90◦, it follows from the nature of exterior angle bisectors that AP0 bisects ∠QAD0. Consequently, the
point P0 has the same distance from AD0 and AQ, equal to the side
length s of square ABCD. But this distance is also the length of the altitude AD in triangle AQP0; then since the altitudes from A and P0 in this triangle are equal, we have AQ = P0Q. Since we can construct P0, we can also construct Q as the intersection of line P X with the perpendicular bisector of the segment AP0. The rest of the construction is obvious, and it is likewise clear that the resulting square ABCD has the required property.
Problem 6 Find all pairs of real numbers a and b such that the system of equations x + y x2+ y2 = a, x3+ y3 x2+ y2 = b
has a solution in real numbers (x, y).
Solution: If the given system has a solution (x, y) for a = A, b = B, then it clearly also has a solution (kx, ky) for a = 1kA, b = kB, for any k 6= 0. It follows that the existence of a solution of the given system depends only on the value of the product ab.
We therefore begin by examining the values of the expression P (u, v) = (u + v)(u
3+ v3)
(u2+ v2)2
where the numbers u and v are normalized by the condition u2+ v2=
1. This condition implies that
P (u, v) = (u + v)(u3+ v3) = (u + v)2(u2− uv + v2)
= (u2+ 2uv + v2)(1 − uv) = (1 + 2uv)(1 − uv).
Under the condition u2 + v2 = 1 the product uv can attain all
values in the interval [−12,12] (if u = cos α and v = sin α, then uv = 12sin 2α). Hence it suffices to find the range of values of the function f (t) = (1 + 2t)(1 − t) on the interval t ∈ [−12,12]. From the formula f (t) = −2t2+ t + 1 = −2 t − 1 4 2 +9 8
it follows that this range of values is the closed interval with endpoints f −1 2 = 0 and f 1 4 = 9 8.
This means that if the given system has a solution, its parameters a and b must satisfy 0 ≤ ab ≤98, where the equality ab = 0 is possible only if x + y = 0 (then, however, a = b = 0).
Conversely, if a and b satisfy 0 < ab ≤ 98, by our proof there exist numbers u and v such that u2+ v2 = 1 and (u + v)(u3+ v3) = ab. Denoting a0 = u + v and b0 = u3+ v3, the equality a0b0 = ab 6= 0
implies that both ratios a0 a and
b
b0 have the same value k 6= 0. But
then (x, y) = (ku, kv) is clearly a solution of the given system for the parameter values a and b.
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1.7
France
Problem 1
(a) What is the maximum volume of a cylinder that is inside a given cone and has the same axis of revolution as the cone? Express your answer in terms of the radius R and height H of the cone. (b) What is the maximum volume of a ball that is inside a given
cone? Again, express your answer in terms of R and H.
(c) Given fixed values for R and H, which of the two maxima you found is bigger?
Solution: Let ` =√R2+ H2 be the slant height of the given cone;
also, orient the cone so that its base is horizontal and its tip is pointing upward.
(a) Intuitively, the cylinder with maximum volume rests against the base of the cone, and the center of the cylinder’s base coincides with the center of the cone’s base. The top face of the cylinder cuts off a smaller cone at the top of the original cone. If the cylinder has radius r, then the smaller cone has height r ·HR and the cylinder has height h = H − r ·HR. Then the volume of the cylinder is πr2h = πr2H1 − r R = 4πR2H r 2R· r 2R· 1 − r R . And by AM-GM on 2Rr , 2Rr , and 1 −Rr this is at most
4πR2H · 1 27 r 2R+ r 2R+ 1 − r R 3 = 4 27πR 2 H, with equality when r/2R = 1 − r/R ⇐⇒ r = 23R.
(b) Intuitively, the sphere with maximum volume is tangent to the base and lateral face of the cone; and its center lies on the cone’s axis. Say the sphere has radius r.
Take a planar cross-section of the cone slicing through its axis; this cuts off a triangle from the cone and a circle from the sphere. The triangle’s side lengths are `, `, and 2R; and its height (from the side of length 2R) is H. The circle has radius r and is the incircle of this triangle.
The area K of the triangle is 12(2R)(H) = RH and its semiperimeter is s = R + `. Then since K = rs we have r = RH
and thus the volume of the sphere is 4 3πr 3=4 3π RH R + ` 3 .
(c) We claim that when h/R = √3 or 2√6, the two volumes are equal; when√3 < h/R < 2, the sphere has larger volume; and when 0 < h/R <√3 or 2 < h/R, the cylinder has larger volume.
We wish to compare 4 27πR 2H and 4 3π RH R+` 3 ; equivalently, multiplying by 4πR272H(R + `)
3, we wish to compare (R + `)3 and
9RH2 = 9R(`2− R2). Writing φ = `/R, this is equivalent to
comparing (1 + φ)3 and 9(φ2− 1). Now,
(1 + φ)3− 9(φ2− 1) = φ3− 6φ2+ 3φ + 10 = (φ + 1)(φ − 2)(φ − 5). Thus when φ = 2 or 5, the volumes are equal; when 2 < φ < 5, the sphere has larger volume; and when 1 < φ < 2 or 5 < φ, the cylinder has larger volume. Comparing R and H instead of R and ` yields the conditions stated before.
Problem 2 Find all integer solutions to (n + 3)n=
n+2
X
k=3
kn. Solution: n = 2 and n = 3 are solutions to the equations; we claim they are the only ones.
First observe that the function f (n) =n+3n+2
n
=1 +n+21
n
is an increasing function for n > 0. To see this, note that the derivative of ln f (n) with respect to n is ln1 +n+21 − n (n+2)(n+3). By the Taylor expansion, ln 1 + 1 n + 2 = ∞ X j=1 1 (n + 2)2j 1 2j − 1(n + 2) − 1 2j >2(n + 2) − 1 2(n + 2)2 and hence d dnln f (n) = ln n + 3 n + 2 − n (n + 2)(n + 3) > 2(n + 2) − 1 2(n + 2)2 − n (n + 2)2 = 3 2(n + 2)2 > 0.
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Now, notice that if f (n) > 2 then we have 2 1 n > 3 2 n > · · · > n + 3 n + 2 n > 2 so that (n + 3)n > 2(n + 2)n > · · · > 2j(n + 3 − j)n> · · · > 2n· (3)n. Then 3n+ 4n+ · · · + (n + 2)n< 1 2n + 1 2n−1+ · · · + 1 2 (n + 3)n = 1 − 1 2n (n + 3)n< (n + 3)n, so the equality does not hold.
Then since 2 < f (6) < f (7) < · · ·, the equality must fail for all n ≥ 6. Quick checks show it also fails for n = 1, 4, 5 (in each case, one side of the equation is odd while the other is even). Therefore the only solutions are n = 2 and n = 3.
Problem 3 For which acute-angled triangle is the ratio of the shortest side to the inradius maximal?
Solution: Let the sides of the triangle have lengths a ≤ b ≤ c; let the angles opposite them be A, B, C; let the semiperimeter be s = 12(a + b + c); and let the inradius be r. Without loss of generality say the triangle has circumradius R = 12and that a = sin A, b = sin B, c = sin C.
The area of the triangle equals both rs = 1
2r(sin A + sin B + sin C)
and abc/4R = 12sin A sin B sin C. Thus r = sin A sin B sin C
sin A + sin B + sin C and
a r =
sin A + sin B + sin C
sin B sin C .
Since A = 180◦−B −C, sin A = sin(B +C) = sin B cos C +cos B sin C and we also have
a
r = cot B + csc B + cot C + csc C.
Note that f (x) = cot x + csc x is a decreasing function along the interval 0◦< x < 90◦. Now there are two cases: B ≤ 60◦, or B > 60◦.
If B ≤ 60◦, then assume that A = B; otherwise the triangle with
angles A0= B0= 12(A + B) ≤ B and C0= C has a larger ratio a0/r0. Then since C < 90◦ we have 45◦< A ≤ 60◦. Now,
a r =
sin A + sin B + sin C
sin B sin C =
2 sin A + sin(2A)
sin A sin(2A) = 2 csc(2A) + csc A. Now csc x has second derivative csc x(csc2x+cot2x), which is strictly positive when 0◦ < x < 180◦; thus both csc x and csc(2x) are strictly convex along the interval 0◦ < x < 90◦. Therefore g(A) =
2 csc(2A) + csc A, a convex function in A, is maximized in the interval 45◦ ≤ A ≤ 60◦ at one of the endpoints. Since g(45◦) = 2 +√2 < 2√3 = g(60◦), it is maximized when A = B = C = 60◦.
As for the case when B > 60◦, since C > B > 60◦, the triangle with A0 = B0 = C0 = 60◦ has a larger ratio a0/r0. Therefore the maximum ratio is 2√3, attained with an equilateral triangle. Problem 4 There are 1999 red candies and 6661 yellow candies on a table, made indistinguishable by their wrappers. A gourmand applies the following algorithm until the candies are gone:
(a) If there are candies left, he takes one at random, notes its color, eats it, and goes to (b).
(b) If there are candies left, he takes one at random, notes its color, and
(i) if it matches the last one eaten, he eats it also and returns to (b).
(ii) if it does not match the last one eaten, he wraps it up again, puts it back, and goes to (a).
Prove that all the candies will eventually be eaten. Find the proba- bility that the last candy eaten is red.
Solution: If there are finitely many candies left at any point, then at the next instant the gourmand must perform either step (a), part (i) of step (b), or part (ii) of step (b). He eats a candy in the first two cases; in the third case, he returns to step (a) and eats a candy. Since there are only finitely many candies, the gourmand must eventually eat all the candies.
We now prove by induction on the total number of candies that if we start with r > 0 red candies and y > 0 yellow candies immediately
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before step (a), then the probability is 12 that the last candy eaten is red.
Suppose that the claim is true for all smaller amounts of candy. After the gourmand first completes steps (a) and (b) exactly once, suppose there are r0 red candies and y0 yellow candies left; we must have r0+ y0< r + y. The chances that r0= 0 is
r r + y · r − 1 r + y − 1· · · · · 1 y + 1= 1 r+y r .
Similarly, the chances that y0 = 0 is 1 (r+y
y )
= 1
(r+y r )
. (In the case r = y = 1, this proves the claim.)
Otherwise, the probability is 1 − 2 (r+y
r )
that both r0 and y0 are still positive. By the induction hypothesis in this case the last candy is equally likely to be red as it is yellow. Thus the overall probability that the last candy eaten is red is
1 r+y r | {z } y0=0 +1 2 1 − 2 r+y r ! | {z } r0,y0>0 = 1 2.
This completes the inductive step, and the proof.
Problem 5 With a given triangle, form three new points by re- flecting each vertex about the opposite side. Show that these three new points are collinear if and only if the the distance between the orthocenter and the circumcenter of the triangle is equal to the diameter of the circumcircle of the triangle.
Solution: Let the given triangle be ABC and let the reflections of A, B, C across the corresponding sides be D, E, F . Let A0, B0, C0 be the midpoints of BC, CA, AB, and as usual let G, H, O denote the triangle’s centroid, orthocenter, and circumcenter. Let triangle A00B00C00 be the triangle for which A, B, C are the midpoints of B00C00, C00A00, A00B00, respectively. Then G is the centroid and H is the circumcenter of triangle A00B00C00. Let D0, E,0F0 denote the projections of O on the lines B00C00, C00A00, A00B00, respectively.
Consider the homothety h with center G and ratio −1/2. It maps A, B, C, A00, B00, C00 into A0, B0, C0, A, B, C, respectively. Note that A0D0 ⊥ BC since O is the orthocenter of triangle A0B0C0. This implies
that h(D) = D0. Similarly, h(E) = E0and h(F ) = F0. Thus, D, E, F
are collinear if and only if D0, E0, F0 are collinear. Now D0, E0, F0 are the projections of O on the sides B00C00, C00A00, A00B00, respectively. By Simson’s theorem, they are collinear if and only if O lies on the circumcircle of triangle A00B00C00. Since the circumradius of triangle A00B00C00 is 2R, O lies on its circumcircle if and only if OH = 2R.
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