The symbols (a,b,\ldots,g) and [a,b,\ldots, g] denote the greatest common divisor and least common multiple, respectively, of the positive integers a,b,\ldots, g. For example, (3,6,18)=3 and [6,15]=30. Prove that
### Solution 1
Consider an arbitrary prime $p$. Let $p^\alpha$, $p^\beta$, and $p^\gamma$ be the greatest powers of $p$ that divide $a$, $b$, and $c$. WLOG let $\alpha \leq \beta \leq \gamma$.
Examining each factor in the equation, we see that the largest power of $p$ that divides the left hand side is $2\gamma -(\beta+\gamma+\gamma) = -\beta$, and the largest power of $p$ that divides the right hand side is 2\alpha -($\alpha + \beta + \alpha) = -\beta$. Since every prime has the same power in both expressions, the expressions are equal.
### Solution 2
Let $p = (a, b, c)$, $pq = (a, b)$, $pr = (b, c)$, and $ps = (c, a)$. Then it follows that $q, r, s$ are pairwise coprime and $a = pqsa'$, $b = pqrb'$, and $c = prsc'$, with $a', b', c'$ pairwise coprime as well. Then, we wish to show $\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} = \frac{p^2}{(pq)(pr)(ps)},$ which can be checked fairly easily.
### Solution 3(very long, but detailed)
Dividing both sides of the equation by $(a,b,c)^2$, and then multiplying both sides by $[a,b][b,c][c,a]$ gives us $\left(\frac{[a,b,c]}{(a,b,c)}\right)^2 = \frac{[a,b][b,c][c,a]}{(a,b)(b,c)(c,a)}.$ Now, we look at the prime factorisations of $a,b,c$. Let $a=2^{x_1}\cdot 3^{x_2}\cdot 5^{x_3}\cdots,$ $b=2^{y_1}\cdot 3^{y_2}\cdot 5^{y_3}\cdots,$ $c=2^{z_1}\cdot 3^{z_2}\cdot 5^{z_3}\cdots,$ where all of the $x_1, y_1, z_1$ are nonnegative integers (they could be 0). Thus, we have $[a,b,c]=2^{\max(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3)} \cdots$ and $(a,b,c)=2^{\min(x_1,y_1,z_1)} \cdot 3^{\min(x_2,y_2,z_2)} \cdot 5^{\min(x_3,y_3,z_3)} \cdots.$ Now, we see that $\left(\frac{[a,b,c]}{(a,b,c)}\right) = 2^{\max(x_1,y_1,z_1) - \min(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2) - \min(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3) - \min(x_3,y_3,z_3)} \cdots.$ Squaring the $\text{LHS}$ will just double all the exponents on the $\text{RHS}$.
Also, we have $[a,b]=2^{\max(x_1,y_1)} \cdot 3^{\max(x_2,y_2)} \cdot 5^{\max(x_3,y_3)} \cdots$. We can also build similar equations for $[b,c],[c,a],(a,b),(b,c)(c,a)$, using 2 of $x,y,z$ and either the minimum of the exponents or the maximum. We see that when we multiply $[a,b]$, $[b,c]$, and $[a,c]$ together, the exponents of the prime factorization will be $\max(x_i,y_i)+ \max(y_i,z_i)+ \max(z_i,x_1)$, where $i$ is chosen for the $i$'th prime. When we multiply $(a,b)$, $(b,c)$, and $(a,c)$ together, the exponents of the prime factorization will be $\min(x_i,y_i)+ \min(y_i,z_i)+ \min(z_i,x_1)$, where $i$ is chosen for the $i$'th prime.
Thus, when we divide the former by the latter, we have $\max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))$. We wish to prove that this is equal to the exponents we got from $([a,b,c]/(a,b,c))^2$.
In other words, this problem is now down to proving that $2(\max(x_i,y_i,z_i) - \min(x_i,y_i,z_i)) = \max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))$ for any nonnegative integers $x_i, y_i, z_i$. WLOG, let $x_i \le y_i \le z_i$. Thus, computing maximums and minimums, this equation turns into $2(z_i - x_i) = y_i + z_i + z_i - x_i - y_i - x_i.
Finally, canceling out the y_i's and collecting like terms, we see the \text{RHS} is 2z_i - 2x_i, which is equal to the \text{LHS}. Thus, this equation is true, and we are done.
Solution 4
Let a=dzxAb=dxyBc=dyzC where \gcd(A,B,C)=\gcd(x,y,z)=1. (Existence proof missing.) Then, \operatorname{lcm}(a,b,c)^2=(dxyzABC)^2 and \operatorname{lcm}(a,b)=dxyzAB. Using this cyclically, we have \text{LHS}=\frac{d^2x^2y^2z^2A^2B^2C^2}{(dxyzAB)(dxyzBC)(dxyzCA)}=\frac{1}{dxyz}. Now, note that \gcd(a,b,c)^2=d^2 and \gcd(a,b)=dx. Using this cyclically, we have \text{RHS}=\frac{d^2}{(dx)(dy)(dz)}=\frac{1}{dxyz}.
Problem 2
A given tetrahedron ABCD is isosceles, that is, AB=CD, AC=BD, AD=BC. Show that the faces of the tetrahedron are acute-angled triangles.
Solution 1
Suppose \triangle ABD is fixed. By the equality conditions, it follows that the maximal possible value of BC occurs when the four vertices are coplanar, with C on the opposite side of \overline{AD} as B. In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.
For the sake of contradiction, suppose \angle ABD is non-acute. Then, (AD)^2\geq (AB)^2+(BD)^2. In our optimal case noted above, ACDB is a parallelogram, so \\2(BD)^2 + 2(AB)^2 = (AD)^2 + (CB)^2 \\ = 2(AD)^2 \\ \geq 2(BD)^2+2(AB)^2.\\ However, as stated, equality cannot be attained, so we get our desired contradiction.
Solution 2
It's not hard to see that the four faces are congruent from SSS Congruence. Without loss of generality, assume that AB\leq BC \leq CA. Now assume, for the sake of contradiction, that each face is non-acute; that is, right or obtuse. Consider triangles \triangle ABC and \triangle ABD. They share side AB. Let k and l be the planes passing through A and B, respectively, that are perpendicular to side AB. We have that triangles ABC and ABD are non-acute, so C and D are not strictly between planes k and l. Therefore the length of CD is at least the distance between the planes, which is AB. However, if CD=AB, then the four points A, B, C, and D are coplanar, and the volume of ABCD would be zero. Therefore CD>AB. However, we were given that CD=AB in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.
Solution 3
Let \vec{a} = \overrightarrow{DA}, \vec{b} = \overrightarrow{DB}, and \vec{c} = \overrightarrow{DC}. The conditions given translate to
We wish to show that \vec{a}\cdot\vec{b}, \vec{b}\cdot\vec{c}, and \vec{c}\cdot\vec{a} are all positive. WLOG, \vec{a}\cdot\vec{a}\geq \vec{b}\cdot\vec{b}, \vec{c}\cdot\vec{c} > 0, so it immediately follows that \vec{a}\cdot\vec{b} and \vec{a}\cdot\vec{c} are positive. Adding all three equations, \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} = 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c}) In addition, \\(\vec{a} - \vec{b} - \vec{c})\cdot(\vec{a} - \vec{b} - \vec{c})\geq 0 \\ \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c}\geq 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} - \vec{b}\cdot\vec{c}) \\ 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c})\geq 2(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} - \vec{b}\cdot\vec{c}) \\ \vec{b}\cdot\vec{c}\geq 0 \\Equality could only occur if \vec{a} = \vec{b} + \vec{c}, which requires the vectors to be coplanar and the original tetrahedron to be degenerate.
Solution 4
Suppose for the sake of contradiction that \angle BAC is not acute. Since all three sides of triangles BAC and CDB are congruent, those two triangles are congruent, meaning \angle BDC=\angle BAC>90^{\circ}. Construct a sphere with diameter BC. Since angles BAC and BDC are both not acute, A and D both lie on or inside the sphere. We seek to make AD=BC to satisfy the conditions of the problem. This can only occur when AD is a diameter of the sphere, since both points lie on or inside the sphere. However, for AD to be a diameter, all four points must be coplanar, as all diameters intersect at the center of the sphere. This would make tetrahedron ABCD degenerate, creating a contradiction. Thus, all angles on a face of an isosceles tetrahedron are acute.
Solution 5
Proof by contradiction: Assume at least one of the tetrahedron's faces are obtuse. WLOG, assume \angle BAC is an obtuse angle. Using SSS congruence to prove that all four faces of the tetrahedron are congruent also shows that the angles surrounding point A are congruent to angles in triangle ABC; namely, \angle BAD is congruent to \angle ABC, and \angle CAD is congruent to \angle ACB. Since the internal angles of triangle ABC must add to 180 degrees, so do the angles surrounding point A. Now lay triangle ABC on a flat surface. A diagram would make it clear that from the perspective of an aerial view, the "apparent" measures of \angle BAD and \angle CAD (which are most likely distorted visually, assuming these angles stick up vertically from the flat surface on which triangle ABC lies) can never exceed the true measures of those angles (equality happens when these angles also lie flat on top of triangle ABC). This means that these two angles can never join to form side AD (because \angle BAC is more than the sum of \angle BAD and \angle CAD - a direct consequence of the facts that \angle BAC is obtuse and all three angles add up to 180 degrees), so the tetrahedron with obtuse triangle faces is impossible.
Solution 6
Lemma: given triangle ABC and the midpoint of BC, which we will call M, we can say that if AM > \frac{BC}{2}, then \angle A < 90. Proof: Since M is the midpoint of BC, BM = MC = \frac{BC}{2}. Since it is given that AM > \frac{BC}{2}, we can substitute \frac{BC}{2} to get two inequalities:AM > CM, \quad AM > BM. The above inequalities imply that \angle C > \angle CAM and \angle B > \angle BAM. Adding these inequalities and simplifying the RHS, we have that \angle C + \angle B > \angle A. Adding \angle A to both sides, replacing the LHS with 180 and dividing by 2 gets us that \angle A < 90. This is our desired inequality, so we are done. Note that all faces of this tetrahedron are congruent, by SSS. In particular, we will use that \triangle ABD \cong \triangle BAC. WLOG, assume that \angle ADB is the largest angle in triangle ABD. Because \triangle ABD \cong \triangle BAC, the median from D to AB is equal length to the median from C to AB. These points meet at E, the midpoint of AB. By the triangle inequality,DE + CE > CD. By substituting CD with AB (this is a given in the problem) and CE with DE, and then dividing by 2, we get that DE > \frac{AB}{2}. By the lemma we showed at the start, this implies that \angle ADB < 90, and since we said that \angle ADB was the largest angle, triangle ADB must be acute. Since all of the faces of this tetrahedron are congruent, then, all of the faces must be acute.
Problem 3
A random number selector can only select one of the nine integers 1, 2, ..., 9, and it makes these selections with equal probability. Determine the probability that after n selections (n>1), the product of the n numbers selected will be divisible by 10.
Solution 1
For the product to be divisible by 10, there must be a factor of 2 and a factor of 5 in there.
The probability that there is no 5 is \left( \frac{8}{9}\right)^n.
The probability that there is no 2 is \left( \frac{5}{9}\right)^n.
The probability that there is neither a 2 nor 5 is \left( \frac{4}{9}\right)^n, which is included in both previous cases.
The only possibility left is getting a 2 and a 5, making the product divisible by 10. By complementarity and principle of inclusion-exclusion, the probability of that is 1- \left( \left( \frac{8}{9}\right)^n + \left( \frac{5}{9}\right)^n - \left( \frac{4}{9}\right)^n\right)=\boxed{1-(8/9)^n-(5/9)^n+(4/9)^n}.
Solution 2(Recursion)
Define a_n as the probability that the product is divisible by 10 after selection n. Likewise, define b_n and c_n with divisibility by 2 and 5, respectively. Define d_n to be the chance of dividing neither 2 nor 5 (and thus not 10 either) similarly.
It is clear that d_n=\left(\frac{4}{9}\right)^n. Now we can define other recursive formulas:
We are able to reach a b state via selecting a non-5 from a b state and selecting an even number from a d state. Thus its formula is b_n=\frac{8}{9}b_{n-1}+\frac{4}{9}d_{n-1}.
We are able to reach a c state via selecting a non-even number from a c state and selecting a 5 from a d state. Thus its formula is c_n=\frac{5}{9}c_{n-1}+\frac{1}{9}d_{n-1}.
Finally, to reach an a state, we can select a 5 from a b state and select an even number from a c state. We can also reach a_n from a_{n-1} because of the fact that once the product is divisible by 10, it will always be divisible by 10 regardless of the following selections. Thus its formula is a_n=a_{n-1}+\frac{1}{9}b_{n-1}+\frac{4}{9}c_{n-1}.
For our formula for b_n, we can substitute to find that b_n=\frac{8}{9}b_{n-1}+\left(\frac{4}{9}\right)^n. Solving this recursion yields b_n=\left(\frac{8}{9}\right)^n-\left(\frac{4}{9}\right)^n.
For our formula for c_n, we can substitute to find that c_n=\frac{5}{9}c_{n-1}+\frac{1}{9}\cdot\left(\frac{4}{9}\right)^{n-1}. Solving this recursion yields c_n=\left(\frac{5}{9}\right)^n-\left(\frac{4}{9}\right)^n.
Finally, substituting the values into the a_n formula yields a_n=a_{n-1}+\frac{1}{9}\left(\left(\frac{8}{9}\right)^{n-1}+4\cdot\left(\frac{5}{9}\right)^{n-1}-5\cdot\left(\frac{4}{9}\right)^{n-1}\right). Solving this recursion yields the final answer, \boxed{1-\left(\frac{8}{9}\right)^n-\left(\frac{5}{9}\right)^n+\left(\frac{4}{9}\right)^n}.
Problem 4
Let R denote a non-negative rational number. Determine a fixed set of integers a,b,c,d,e,f, such that for every choice of R,
We cross multiply to get a\sqrt[3]{4}+b\sqrt[3]{2}+c=2d+e\sqrt[3]{4}+f\sqrt[3]{2}. It's not hard to show that, since a, b, c, d, e, and f are integers, then a=e, b=f, and c=2d.
Note, however, that this is a necessary but insufficient condition. For example, we must also have a^2<2bc to ensure the function does not have any vertical asymptotes (which would violate the desired property). A simple search shows that a=0, b=2, and c=2 works.
Problem5
A given convex pentagon ABCDE has the property that the area of each of the five triangles ABC, BCD, CDE, DEA, and EAB is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property.
Solution 1
Lemma: Convex pentagon A_0A_1A_2A_3A_4 has the property that [A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1] if and only if \overline{A_{n - 1}A_{n + 1}}\parallel\overline{A_{n - 2}A_{n + 2}} for n = 0, 1, 2, 3, 4 (indices taken mod 5).
Proof: For the "only if" direction, since [A_0A_1A_2] = [A_1A_2A_3], A_0 and A_3 are equidistant from \overline{A_1A_2}, and since the pentagon is convex, \overline{A_0A_3}\parallel\overline{A_1A_2}. The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed.
Let A'B'C'D'E' be the inner pentagon, labeled so that A and A' are opposite each other, and let a, b, c, d, e be the side lengths of the inner pentagon labeled opposite their corresponding vertices. From the lemma, pentagon A'B'C'D'E' is similar to pentagon ABCDE with AB = m(A'B') and parallelogram ABCB' (and cyclic) has area 2. Supposing \\ P = [ABCDE] \\ Q = [A'B'C'D'E'] \\ R = \sum_{\text{cyc}}[AC'D'] \\ S = \sum_{\text{cyc}}[ABD'],\\ we have \\ \sum_{\text{cyc}}[ABCB'] = 5Q + 3R + 2S \\ 10 = 3Q + R + 2P.\\ Since BCDC' and CDED' are parallelograms, BD' = EC' = (m - 1)a. Triangle EB'C' is similar to triangle ECB, so me = BC = \left(\frac{2m - 1}{m - 1}\right)e, and with the requirement that m > 1,m = \frac{2m - 1}{m - 1}\Rightarrow m = \frac{3 + \sqrt{5}}{2}. Now, we compute that [AC'D'] = \frac{1}{2m - 1} = \sqrt{5} - 2, and similar computation for the other four triangles gives R = 5\sqrt{5} - 10. From the aforementioned pentagon similarity, Q = P/m^2 = \left(\frac{7 - 3\sqrt{5}}{2}\right)P. Solving for P, we have \\ 10 = 3\left(\frac{7 - 3\sqrt{5}}{2}\right)P + (5\sqrt{5} - 10) + 2P \\ 20 - 5\sqrt{5} = \left(\frac{25 - 9\sqrt{5}}{2}\right)P \\ P = \frac{40 - 10\sqrt{5}}{25 - 9\sqrt{5}} \\ = \frac{5 + \sqrt{5}}{2}.
To show that there are infinitely many pentagons with the given property, we start with triangle A'B'C' and construct the original pentagon using the required length ratios. This triangle has three continuous degrees of freedom, and the condition that the equal areas be specifically unity means that there are two continuous degrees of freedom in which the configuration can vary, which implies the desired result.
Alternatively, one can shear a regular pentagon, a transformation which preserves areas and the property of a pair of lines being parallel.