3. DIAGNOSTICO AMBIENTAL - LINEA BASE
3.1.2. MEDIO BIÓTICO
Problem 7.1 Determine the side lengths of a right triangle if they are integers and the product of the leg lengths is equal to three times the perimeter.
Solution: One of the leg lengths must be divisible by 3; let the legs have lengths 3a and b and let the hypotenuse have length c, where a, b, and c are positive integers. From the given condition we have 3ab = 3(3a + b + c), or c = ab − 3a − b. By the Pythagorean theorem, we have (3a)2+ b2= c2= (ab − 3a − b)2, which simplifies to
ab[(a − 2)(b − 6) − 6] = 0.
Since a, b > 0, we have (a, b) ∈ {(3, 12), (4, 9), (5, 8), (8, 7)}, and there- fore the side lengths of the triangle are either (9, 12, 15), (8, 15, 17), or (7, 24, 25).
Problem 7.2 Let a, b, c be nonzero integers, a 6= c, such that a
c =
a2+ b2
c2+ b2.
Prove that a2+ b2+ c2 cannot be a prime number.
Solution: Cross-multiplying and factoring, we have (a − c)(b2−
ac) = 0. Since a 6= c, we have ac = b2. Now, a2 + b2 + c2 =
a2+ (2ac − b2) + c2= (a + c)2− b2= (a + b + c)(a − b + c). Also, |a|, |c|
cannot both be 1. Then a2+b2+c2> |a|+|b|+|c| ≥ |a+b+c|, |a−b+c|,
whence a2+ b2+ c2cannot be a prime number.
Problem 7.3 Let ABCD be a convex quadrilateral with ∠BAC = ∠CAD and ∠ABC = ∠ACD. Rays AD and BC meet at E and rays AB and DC meet at F . Prove that
(a) AB · DE = BC · CE; (b) AC2< 1
2(AD · AF + AB · AE).
Solution:
(a) Because ∠BAC + ∠CBA = ∠ECA, we have ∠ECD = ∠BAC. Then 4CDE ∼ 4ACE, and CEDE = AECE. But since AC is the
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angle bisector of ∠A in triangle ABE, we also have AECE = AB BC.
Thus CEDE = ABBC, whence AB · DE = BC · CE.
(b) Note that AC is an angle bisector of both triangle ADF and triangle AEB. Thus it is enough to prove that if XL is an angle bisector in an arbitrary triangle XY Z, then XL2 < XY · XZ.
Let M be the intersection of−XL and the circumcircle of triangle−→ XY Z. Because 4XY L ∼ 4XM Z, we have XL2< XL · XM =
XY · XZ, as desired.
Problem 7.4 In triangle ABC, D and E lie on sides BC and AB, respectively, F lies on side AC such that EF k BC, G lies on side BC such that EG k AD. Let M and N be the midpoints of AD and BC, respectively. Prove that
(a) EF
BC +
EG
AD = 1;
(b) the midpoint of F G lies on line M N .
Solution:
(a) Since EF k BC, 4AEF ∼ 4ABC and EFBC = AEAB. Similarly,
since EG k AD, 4BEG ∼ 4BAD and ADEG = EBAB. Hence
EF BC +
EG AD = 1.
(b) Let lines AN, EF intersect at point P , and let Q be the point on line BC such that P Q k AD. Since BC k EF and N is the midpoint of BC, P is the midpoint of EF . Then vector EP equals both vectors P F and GQ, and P F QG is a parallelogram. Thus the midpoint X of F G must also be the midpoint of P Q. But then since M is the midpoint of AD and AD k P Q, points M, X, N must be collinear.
Problem 8.1 Let p(x) = 2x3− 3x2+ 2, and let
S = {p(n) | n ∈ N, n ≤ 1999}, T = {n2+ 1 | n ∈ N},
U = {n2+ 2 | n ∈ N}.
Prove that S ∩ T and S ∩ U have the same number of elements. Solution: Note that |S ∩ T | is the number of squares of the form 2n3− 3n2+ 1 = (n − 1)2
n ≤ 1999, (n − 1)2(2n + 1) is a square precisely when either n = 1 or
when n ∈ {12(k2− 1) | k = 1, 3, 5, . . . , 63}. Thus, |S ∩ T | = 33.
Next, |S ∩ U | is the number of squares of the form 2n3− 3n2 =
n2(2n − 3) where n ∈ N, n ≤ 1999. And for n ≤ 1999, n2(2n − 3) is a square precisely when either n = 0 or when n ∈ {12(k2+ 3) | k =
1, 3, 5, . . . , 63}. Thus |S ∩ U | = 33 as well, and we are done. Problem 8.2
(a) Let n ≥ 2 be a positive integer and
x1, y1, x2, y2, . . . , xn, yn
be positive real numbers such that
x1+ x2+ · · · + xn ≥ x1y1+ x2y2+ · · · + xnyn. Prove that x1+ x2+ · · · + xn≤ x1 y1 +x2 y2 + · · · +xn yn . (b) Let a, b, c be positive real numbers such that
ab + bc + ca ≤ 3abc. Prove that
a3+ b3+ c3≥ a + b + c.
Solution:
(a) Applying the Cauchy-Schwarz inequality and then the given inequality, we have n X i=1 xi !2 ≤ n X i=1 xiyi· n X i=1 xi yi ≤ n X i=1 xi· n X i=1 xi yi . Dividing both sides byPn
i=1xi yields the desired inequality.
(b) By the AM-HM inequality on a, b, c we have
a + b + c ≥ 9abc
ab + bc + ca ≥ 9abc 3abc = 3.
Then, since the given condition is equivalent to 1a+1b+1c ≤ 3, we have a + b + c ≥ 1a+1b +1c. Hence setting x1= a, x2= b, x3= c
and y1 = a12, y2 = b12, y3 = c12 in the result from part (a) gives
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Problem 8.3 Let ABCDA0B0C0D0 be a rectangular box, let E
and F be the feet of perpendiculars from A to lines A0D and A0C respectively, and let P and Q be the feet of perpendiculars from B0 to lines A0C0 and A0C respectively. Prove that
(a) planes AEF and B0P Q are parallel; (b) triangles AEF and B0P Q are similar.
Solution:
(a) Let (P1P2. . . Pk) denote the plane containing points P1, P2, . . . , Pk.
First observe that quadrilateral A0B0CD is a parallelogram and thus lies in a single plane.
We are given that AE ⊥ A0D. Also, line AE is contained in plane (ADD0A), which is perpendicular to line CD. Hence AE ⊥ CD as well, and therefore AE ⊥ (A0B0CD) and AE ⊥ A0C. And since we know that A0C ⊥ AF, we have A0C ⊥ (AEF ) and A0C ⊥ EF.
Likewise, B0Q ⊥ A0C. And since lines EF, B0Q, and A0C all lie in plane (A0B0CD), it follows that EF k B0Q. In a similar way we deduce that AF k P Q. Hence the planes (AEF ) and (B0P Q) are parallel, as desired.
(b) Since EF k B0Q and F A k QP, we have ∠EF A = ∠P QB0.
Furthermore, from above AE ⊥ EF and likewise B0P ⊥ P Q, implying that ∠AEF = ∠B0P Q = 90◦ as well. Therefore 4AEF ∼ 4B0P Q, as desired.
Problem 8.4 Let SABC be a right pyramid with equilateral base ABC, let O be the center of ABC, and let M be the midpoint of BC. If AM = 2SO and N is a point on edge SA such that SA = 25SN , prove that planes ABP and SBC are perpendicular, where P is the intersection of lines SO and M N .
Solution: Let AB = BC = CA = s. Then some quick calculations show that AO = √ 3 3 s, AM = √ 3 2 s, AS = 5 √ 48s, and AN = 24 5√48s.
Then AO · AM = AN · AS = 12s2, whence M ON S is a cyclic
quadrilateral. Thus, ∠M N S = 90◦, and P is the orthocenter of triangle AM S. Let Q be the intersection of lines AP and M S.
Note that ∠AM B = ∠AQM = ∠QM B = 90◦. From repeated
M B2= AQ2+ QM2+ M B2= AQ2+ QB2
, whence ∠AQB = 90◦. Now AQ ⊥ QB and AQ ⊥ QM , so line AQ must be perpendicular to plane SBC. Then since plane ABP contains line AQ, planes ABP and SBC must be perpendicular.
Problem 9.1 Let ABC be a triangle with angle bisector AD. One considers the points M, N on rays AB and AC respectively, such that ∠M DA = ∠ABC and ∠N DA = ∠BCA. Lines AD and M N meet at P . Prove that
AD3= AB · AC · AP.
Solution: Since 4ADB ∼ 4AM D, AD
AB = AM
AD. Also, ∠M AN +
∠N DM = π, whence AM DN is cyclic. Since ∠DCA = ∠ADN = ∠AM N , 4ADC ∼ 4AP M , and ADAP =
AC AM. Therefore, AD AB AD AC AD AP = AM AD AD AC AC AM = 1.
Problem 9.2 For a, b > 0, denote by t(a, b) the positive root of the equation
(a + b)x2− 2(ab − 1)x − (a + b) = 0.
Let M = {(a, b) | a 6= b, t(a, b) ≤√ab}. Determine, for (a, b) ∈ M , the minimum value of t(a, b).
Solution: Consider the polynomial P (x) = (a + b)x2− 2(ab − 1)x −
(a + b) = 0. Since a + b 6= 0, the product of its roots is −a+ba+b = −1. Hence P must have a unique positive root t(a, b) and a unique negative root. Since the leading coefficient of P (x) is positive, the graph of P (x) is positive for x > t(a, b) and negative for 0 ≤ x < t(a, b) (since in the latter case, x is between the two roots). Thus, the condition t(a, b) ≤√ab is equivalent to P (√ab) ≥ 0, or
(ab − 1)(a + b − 2√ab) ≥ 0.
But a + b > 2√ab by AM-GM, where the inequality is sharp since a 6= b. Thus t(a, b) ≤√ab exactly when ab ≥ 1.
Now using the quadratic formula, we find that
t(a, b) = ab − 1 a + b + s ab − 1 a + b 2 + 1.
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Problem 9.3 In the convex quadrilateral ABCD the bisectors of angles A and C meet at I. Prove that there exists a circle inscribed in ABCD if and only if
[AIB] + [CID] = [AID] + [BIC].
Solution: It is well known that a circle can be inscribed in a convex quadrilateral ABCD if and only if AB + CD = AD + BC. The bisector of angle A consists of those points lying inside ∠BAD equidistant from lines AB and AD; similarly, the bisector of angle C consists of those points lying inside ∠BCD equidistant from lines BC and BD.
Suppose ABCD has an incircle. Then its center is equidistant from all four sides of the quadrilateral, so it lies on both bisectors and hence equals I. If we let r denote the radius of the incircle, then we have
[AIB] + [CID] = r(AB + CD) = r(AD + BC) = [AID] + [BIC]. Conversely, suppose that [AIB] + [CID] = [AID] + [BIC]. Let d(I, `) denote the distance from I to any line `, and write x = d(I, AB) = d(I, AD) and y = d(I, BC) = d(I, CD). Then
[AIB] + [CID] = [AID] + [BIC] AB · x + CD · y = AD · x + BC · y
x (AB − AD) = y (BC − CD).
If AB = AD, then BC = CD and it follows that AB + CD = AD + BC. Otherwise, suppose that AB > AD; then BC > CD as well. Consider the points A0 ∈ AB and C0 ∈ BC such that
AD = AA0 and CD = CC0. By SAS, we have 4AIA0∼= 4AID and 4DCI ∼= 4C0IC. Hence IA0= ID = IC0. Furthermore, subtracting [AIA0] + [DCI] = [AID] + [C0IC] from both sides of our given condition, we have [A0IB] = [C0IB] or IA0· IB · sin ∠A0IB = IC0·
IB · sin ∠CIB. Thus ∠A0IB = ∠C0IB, and hence 4A0IB ∼= C0IB by SAS.
Thus ∠IBA0 = ∠IBC0, implying that I lies on the angle bisector
of ∠ABC. Therefore x = d(I, AB) = d(I, BC) = y, and the circle centered at I with radius x = y is tangent to all four sides of the quadrilateral.
(a) Let a, b ∈ R, a < b. Prove that a < x < b if and only if there exists 0 < λ < 1 such that x = λa + (1 − λ)b.
(b) The function f : R → R has the property:
f (λx + (1 − λ)y) < λf (x) + (1 − λ)f (y)
for all x, y ∈ R, x 6= y, and all 0 < λ < 1. Prove that one cannot find four points on the function’s graph that are the vertices of a parallelogram.
Solution:
(a) No matter what x is, there is a unique value λ = b−xb−a such that x = λa + (1 − λ)b; and 0 < b−xb−a < 1 ⇐⇒ a < x < b, which proves the claim.
(b) The condition is Jensen’s inequality and shows that the function f is strictly convex. Stated geometrically, whenever x < t < y the point (t, f (t)) lies strictly below the line joining (x, f (x)) and (y, f (y)). Suppose there were a parallelogram on the graph of f whose vertices, from left to right, have x-coordinates a, b, d, c. Then either (b, f (d)) or (d, f (d)) must lie on or above the line joining (a, f (a)) and (c, f (c)), a contradiction.
Problem 10.1 Find all real numbers x and y satisfying 1 4x+ 1 27y = 5 6 log27y − log4x ≥ 1 6 27y− 4x≤ 1.
Solution: First, for the second equation to make sense we must have x, y > 0 and thus 27y > 1. Now from the third equation we
have
1 27y ≥
1 4x+ 1,
which combined with the first equation gives 1 4x + 1 4x+ 1 ≤ 5 6,
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whence x ≥ 12. Similarly, the first and third equations also give 5 6 ≤ 1 27y− 1 + 1 27y,
whence y ≤ 13. If either x > 12 or y < 13, we would have log27y −
log4x < 16, contradicting the second given equation. Thus, the only solution is (x, y) = 1
2, 1
3, which indeed satisfies all three equations.
Problem 10.2 A plane intersects edges AB, BC, CD, DA of the regular tetrahedron ABCD at points M, N, P, Q, respectively. Prove that
M N · N P · P Q · QM ≥ AM · BN · CP · DQ.
Solution: By the law of cosines in triangle M BN , we have M N2= M B2+ BN2− M B · BN ≥ M B · BN.
Similarly, N P2≥ CN · CP, P N2≥ DP · DQ, and M Q2≥ AQ · AM.
Multiplying these inequalities yields (M N · N P · P Q · M Q)2≥
(BM · CN · DP · AQ) · (AM · BN · CP · DQ) .
Now the given plane is different from plane (ABC) and (ADC). Thus if it intersects line AC at some point T, then points M, N, T must be collinear—because otherwise, the only plane containing M, N, T would be plane (ABC). Therefore it intersects line AC at most one point T, and by Menelaus’ Theorem applied to triangle ABC and line M N T we have
AM · BN · CT M B · N C · T A = 1. Similarly, P, Q, T are collinear and
AQ · DP · CT QD · P C · T A= 1.
Equating these two fractions and cross-multiplying, we find that AM · BN · CP · DQ = BM · CN · DP · AQ.
This is true even if the plane does not actually intersect line AC: in this case, we must have M N k AC and P Q k AC, in which case
ratios of similar triangles show that AM · BN = BM · CN and CP · DQ = DP · AQ.
Combining this last equality with the inequality from the first paragraph, we find that
(M N · N P · P Q · QM )2≥ (AM · BN · CP · DQ)2, which implies the desired result.
Problem 10.3 Let a, b, c (a 6= 0) be complex numbers. Let z1and
z2 be the roots of the equation az2+ bz + c = 0, and let w1 and w2
be the roots of the equation
(a + c)z2+ (b + b)z + (a + c) = 0. Prove that if |z1|, |z2| < 1, then |w1| = |w2| = 1.
Solution: We begin by proving that Re (b)2≤ |a + c|2. If z
1= z2=
0, then b = 0 and the claim is obvious. Otherwise, write a = m + ni and c = r + si; and write z1= x + yi where t = |z1| =
p
x2+ y2< 1.
Also note that
r2+ s2= |c|2= |az1z2|2< |a|2|z1|2= (m2+ n2)t2. (1)
Assume WLOG that z1 6= 0. Then | Re (b)| = | Re (−b)| =
| Re (az1+ c/z1)| = | Re (az1) + Re (c/z1)|; that is,
| Re (b)| = |(mx − ny) + (rx + sy)/t2|
= |x(m + r/t2) + y(s/t2− n)| ≤px2+ y2p
(m + r/t2)2+ (s/t2− n)2
= tp(m + r/t2)2+ (s/t2− n)2,
where the inequality follows from Cauchy-Schwarz. Proving our claim then reduces to showing that
t2 (m + r/t2)2+ (s/t2− n)2 ≤ (m + r)2+ (n − s)2
⇐⇒ (mt2+ r)2+ (st2− n)2≤ t2 (m + r)2+ (n − s)2 ⇐⇒ (r2+ s2)(1 − t2) < (m2+ n2)(t4− t2)
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But 1 − t2> 0 by assumption, and (m2+ n2)t2− (r2+ s2) > 0 from
(1); therefore our claim is true.
Now since |c/a| = |z1z2| < 1, we have |c| < |a| and a + c 6= 0. Then
by the quadratic equation, the roots to (a+c)z2+(b+b)z +(a+c) = 0
are given by
−(b + b) ±q(b + b)2− 4(a + c)(a + c)
2(a + c) ,
or (dividing the numerator and denominator by 2) − Re (b) ±p Re (b)2− |a + c|2
a + c =
− Re (b) ± ip|a + c|2− Re (b)2
a + c .
When evaluating either root, the absolute value of the numerator is p Re (b)2+ (|a + c|2− Re (b)2) = |a + c|; and the absolute value of
the denominator is clearly |a + c| as well. Therefore indeed |w1| =
|w2| = 1, as desired.
Problem 10.4
(a) Let x1, y1, x2, y2, . . . , xn, yn be positive real numbers such that
(i) x1y1< x2y2< · · · < xnyn;
(ii) x1+ x2+ · · · + xk ≥ y1+ y2+ · · · + yk for all k = 1, 2, . . . , n.
Prove that 1 x1 + 1 x2 + · · · + 1 xn ≤ 1 y1 + 1 y2 + · · · + 1 yn .
(b) Let A = {a1, a2, . . . , an} ⊂ N be a set such that for all distinct
subsets B, C ⊆ A, X x∈B x 6= X x∈C x. Prove that 1 a1 + 1 a2 + · · · + 1 an < 2. Solution: (a) Let πi = x1
iyi, δi = xi− yi for all 1 ≤ i ≤ n. We are given that
π1 > π2> · · · > πn > 0 and thatPki=1δi≥ 0 for all 1 ≤ k ≤ n.
Note that n X k=1 1 yk − 1 xk = n X k=1 πkδk
= πn n X i=1 δi+ n−1 X k=1 (πk− πk+1) (δ1+ δ2+ · · · + δk) ≥ 0, as desired.
(b) Assume without loss of generality that a1 < a2 < · · · < an, and
let yi = 2i−1for all i. Clearly,
a1y1< a2y2< · · · < anyn.
For any k, the 2k− 1 sums made by choosing at least one of the
numbers a1, a2, . . . , ak are all distinct. Hence the largest of them,
Pk
i=1ai, must be at least 2
k− 1. Thus for all k = 1, 2, . . . , n we
have
a1+ a2+ · · · + ak ≥ 2k− 1 = y1+ y2+ · · · + yk.
Then by part (a), we must have 1 a1 + 1 a2 + · · · + 1 an < 1 y1 + 1 y2 + · · · + 1 yn = 2 − 1 2n−1 < 2, as desired.