3rd All Russian Mathematical Olympiad Problems 1963
1. Given 5 circles. Every 4 have a common point. Prove that there is a point common to all 5.
2. 8 players compete in a tournament. Everyone plays everyone else just once. The winner of a game gets 1, the loser 0, or each gets 1/2 if the game is drawn. The final result is that everyone gets a different score and the player placing second gets the same as the total of the four bottom players. What was the result of the game between the player placing third and the player placing seventh?
3. (a) The two diagonals of a quadrilateral each divide it into two parts of equal area. Prove it is a parallelogram.
(b) The three main diagonals of a hexagon each divide it into two parts of equal area. Prove they have a common point. [If ABCDEF is a hexagon, then the main diagonals are AD, BE and CF.]
4. The natural numbers m and n are relatively prime. Prove that the greatest common divisor of m+n and m2+n2 is either 1 or 2.
5. Given a circle c and two fixed points A, B on it. M is another point on c, and K is the midpoint of BM. P is the foot of the perpendicular from K to AM.
(a) prove that KP passes through a fixed point (as M varies);
(b) find the locus of P.
6. Find the smallest value x such that, given any point inside an equilateral triangle of side 1, we can always choose two points on the sides of the triangle, collinear with the given point and a distance x apart.
7. (a) A 6 x 6 board is tiled with 2 x 1 dominos. Prove that we can always divide the board into two rectangles each of which is tiled separately (with no domino crossing the dividing line).
(b) Is this true for an 8 x 8 board?
8. Given a set of n different positive reals {a1, a2, ... , an}. Take all possible non-empty subsets and form their sums. Prove we get at least n(n+1)/2 different sums.
9. Given a triangle ABC. Let the line through C parallel to the angle bisector of B meet the angle bisector of A at D, and let the line through C parallel to the angle bisector of A meet the angle bisector of B at E. Prove that if DE is parallel to AB, then CA=CB.
10. An infinite arithmetic progression contains a square. Prove it contains infinitely many squares.
11. Can we label each vertex of a 45-gon with one of the digits 0, 1, ... , 9 so that for each pair of distinct digits i, j one of the 45 sides has vertices labeled i, j?
12. Find all real p, q, a, b such that we have (2x-1)20 - (ax+b)20 = (x2+px+q)10 for all x.
13. We place labeled points on a circle as follows. At step 1, take two points at opposite ends of a diameter and label them both 1. At step n>1, place a point at the midpoint of each arc created at step n-1 and label it with the sum of the labels at the two adjacent points. What is the total sum of the labels after step n?
For example, after step 4 we have: 1, 4, 3, 5, 2, 5, 3, 4, 1, 4, 3, 5, 2, 5, 3, 4.
14. Given an isosceles triangle, find the locus of the point P inside the triangle such that the distance from P to the base equals the geometric mean of the distances to the sides.
Solutions
Problem 1
Given 5 circles. Every 4 have a common point. Prove that there is a point common to all 5.
Solution
Let the circles be a, b, c, d, e. Let A be a point common to b, c, d, e, let B be a point common to a, c, d, e and so on. If any two of A, B, C, D, E coincide then the coincident point is on all 5 circles. Suppose they are all distinct. Then A, B, C are on d and e. Hence d and e coincide (3 points determine a circle). Hence D is on all 5 circles.
Problem 2
8 players compete in a tournament. Everyone plays everyone else just once. The winner of a game gets 1, the loser 0, or each gets 1/2 if the game is drawn. The final result is that everyone gets a different score and the player placing second gets the same as the total of the four bottom players. What was the result of the game between the player placing third and the player placing seventh?
Solution
The bottom 4 played 6 games amongst themselves, so their scores must total at least 6. Hence the number 2 player scored at least 6. The maximum score possible is 7, so if the number 2 player scored more than 6, then he must have scored 6 1/2 and the top player 7. But then the top player must have won all his games, and hence the number 2 player lost at least one game and could not have scored 6 1/2. Hence the number 2 player scored exactly 6, and the bottom 4 players lost all their games with the top 4 players. In particular, the number 3 player won against the number 7 player.
Problem 3
(a) The two diagonals of a quadrilateral each divide it into two parts of equal area. Prove it is a parallelogram.
(b) The three main diagonals of a hexagon each divide it into two parts of equal area. Prove they have a common point. [If ABCDEF is a hexagon, then the main diagonals are AD, BE and CF.]
Solution
(a) Let the quadrilateral be ABCD and let the diagonals AC, BD meet at E. Then area ABC = AC.EB.sin CEB/2, and area ADC = AC.ED.sin CEB/2, so E is the midpoint of BD. Similarly, it is the midpoint of AC. Hence the triangles AEB and CED are congruent, so angle CDE = angle ABE, and hence AB is parallel to CD. Similarly, AD is parallel to BC.
(b) Let the hexagon be ABCDEF. Let BE, CF meet at J, let AD, CF meet at K, and let AD, BE meet at L. Let AK=a, BJ=b, CJ=c, DL=d, EL=e, FK=f. Also let KL=x, JL=y and JK=z. Consider the pair of diagonals AD, BE. They divide the hexagon into 4 parts: the triangles ALB and DLE, and the quadrilaterals AFEL and BCDL. Since area ALB + area AFEL = area DLE + area BCDL, and area ALB + area BCDL = area DLE + area AFEL, the two triangles must have the same area (add the two inequalities). But area ALB = 1/2 AL.BL.sin ALB, and area DLE = 1/2 DL.EL.sin DLE = 1/2 DL.EL.sin ALB, so AL.BL = DL.EL or de = (a+x)(b+y). Similarly, considering the other two paris of diagonals, we get bc = (e+y)(f+z) and af = (c+z)(d+x). Multiplying the three inequalities gives: abcdef = (a+f)(b+y)(c+z)(d+x)(e+y)(f+z). But x, y, z are non-negative, so they must be zero and hence the three diagonals pass through a common point.
Problem 4
The natural numbers m and n are relatively prime. Prove that the greatest common divisor of m+n and m2+n2 is either 1 or 2.Solution
If d divides m+n and m2+n2, then it also divides (m+n)2 - (m2+n2) = 2mn and hence also 2m(m+n) - 2mn = 2m2 and 2n(m+n) - 2mn = 2n2. But m and n are relatively prime, so m2 and n2 are also. Hence d must divide 2.
Problem 5
Given a circle c and two fixed points A, B on it. M is another point on c, and K is the midpoint of BM. P is the foot of the perpendicular from K to AM.
(a) prove that KP passes through a fixed point (as M varies);
(b) find the locus of P.
Solution
(a) Take Y on the circle so that angle ABY=90. Then AY is a diameter and so angle AMY=90. Take X as the midpoint of BY. Then triangles BXK and BYM are similar, so XK is parallel to YM. Hence XK is perpendicular to AM, and so P is the intersection of XK and AM. In other words, KP always passes through X.
(b) P must lie on the circle diameter AX, and indeed all such points can be obtained (given a point P on the circle, take M as the intersection of AP and the original circle). So the locus of P is the circle diameter AX.
Problem 6
Find the smallest value x such that, given any point inside an equilateral triangle of side 1, we can always choose two points on the sides of the triangle, collinear with the given point and a distance x apart.
Solution
Answer: 2/3.
Let O be the center of ABC. Let AO meet BC at D, let BO meet CA at E, and let CO meet AB at F. Given any point X inside ABC, it lies in one of the quadrilaterals AEOF, CDOE, BFOD. Without loss of generality, it lies in AEOF. Take the line through X parallel to BC. It meets AB in P and AC in Q. Then PQ is shorter than the parallel line MON with M on AB and N on AC, which has length 2/3. If we twist the segment PXQ so that it continues to pass through X, and P remains on AB and Q on AC, then its length will change continuously. Eventually, one end will reach a vertex, whilst the other will be on the opposite side and hence the length of the segment will be at least that of an altitude, which is greater than 2/3. So at some intermediate position its length will be 2/3.
To show that no value smaller than 2/3 is possible, it is sufficient to show that any segment POQ with P and Q on the sides of the triangle has length at least 2/3. Take P on MB and Q on AN with P, O, Q collinear. Then PQ cos POM = MN - QN cospi/3 + PM cos pi/3. But PM>QN (using the sine rule, PM = OM sinPOM/sinOPM and QN = ON sinQON/sinOQN, but OM=ON, angle POM = angle QON, and angle OQN = angle OPM + pi/3 > angle OPM), and hence PQ > MN sec POM > MN.
Problem 7
(a) A 6 x 6 board is tiled with 2 x 1 dominos. Prove that we can always divide the board into two rectangles each of which is tiled separately (with no domino crossing the dividing line).
#(b) Is this true for an 8 x 8 board?
Solution
(a) We say a domino bridges two columns if half the domino is in each column. We show that for 0<n<6 the number of dominoes bridging columns n and n+1 must be at least 2 and even.
Consider first n=1. There cannot be 3 dominoes entirely in column 1, or it would be separately tiled. So there must be at least one domino bridging columns 1 and 2. The number must be even, because it must equal the number of squares in column 1 (even) less twice the number of dominoes (entirely) in column 1.
Now suppose it is true for n<5 and consider column n+1. There must be at least one domino bridging columns n+1 and n+2, or columns 1 thru n+1 would be separately tiled. The number must be even, because it must equal the number of squares in column n+1 (even) less the number bridging n and n+1 (even) less twice the number entirely in the column.
So in total there are at least 5 x 2 = 10 dominoes bridging columns. By the same argument there are at least another 10 bridging rows, but there are only 18 dominoes in total.
(b) No. For example:
Problem 8
Given a set of n different positive reals {a1, a2, ... , an}. Take all possible non-empty subsets and form their sums. Prove we get at least n(n+1)/2 different sums.
Solution
Assume a1 < a2 < ... < an. We have the following collection of increasing sums:
A total of 1+2+ ... +n = n(n+1)/2.
Problem 9
Given a triangle ABC. Let the line through C parallel to the angle bisector of B meet the angle bisector of A at D, and let the line through C parallel to the angle bisector of A meet the angle bisector of B at E. Prove that if DE is parallel to AB, then CA=CB.
Solution
The idea is to find an expression for the perpendicular distance h from D to AB. Let γ = ∠ACB, α = ½∠CAB, and β = ½∠ABC. We have h = AP sin α.
Using the sine rule on APC, we have AP = AC sin(γ+β)/sin(α+β), so h = AC sin α sin(γ+β)/sin(α+β). Similarly, the perpendicular distance k from E to AB is BC sin β sin(γ+α)/sin(α+β).
We also have that AC/BC = sin 2β/sin 2α, and hence h/k = sin 2β sin α sin(γ+β)/(sin 2α sin β sin(γ+α)). Using the fact that sin(γ+β) = sin(2α+β), and the expression for sin 2θ, we get h/k = (sin(2α+2β) + sin 2α)/(sin(2α+2β) + sin 2β) and hence h = k iff the triangle is isosceles.
For some reason the geometric solution took me longer to find. Let ED meet BC at X. Then XCD and XBE are isosceles, so BC = BX + XC = DX + XE = DE. Similarly, AC = DE. Hence AC = BC.
Problem 10
An infinite arithmetic progression contains a square. Prove it contains infinitely many squares.
Solution
Let the square be a2 and the difference d, so that all numbers of the form a2+nd belong to the arithmetic progression (for n a natural number). Take n to be 2ar + dr2, then a2+nd = (a+dr)2.
Problem 11
Can we label each vertex of a 45-gon with one of the digits 0, 1, ... , 9 so that for each pair of distinct digits i, j one of the 45 sides has vertices labeled i, j?
Solution
10 x 5 > 45, so some digit i0 must appear less than 5 times. But each occurrence can give at most 2 edges i0, j, so there are at most 8 edges i0, j, which is one too few.
Problem 12
Find all real p, q, a, b such that we have (2x-1)20 - (ax+b)20 = (x2+px+q)10 for all x.
Solution
Comparing coefficients of x20, we must have a = (220 - 1)1/20 (note that we allow either the positive or the negative root).
Set x=1/2. Then we must have (ax + b)20 = 0 = (x2+px+q)10, and hence ax + b =0 and x2+px+q = 0. So b = -a/2, and 1/4 + p/2 + q = 0.
Set x=0. Then we get q10 = 1 - b20 = 1/220, so q = 1/4 or -1/4, and p= -1 or 0 respectively. Comparing the coefficients of x19, we must have p = -1 and q = 1/4. So, if there is a solution, then it must be: a = (220 - 1)1/20, b = -a/2, p = -1, q = 1/4. This is indeed a solution because with these values, the lhs = 220(x - 1/2)20 - (x - 1/2)20a20 = (x - 1/2)20 = (x2 - x + 1/4)10 = rhs.
Problem 13
We place labeled points on a circle as follows. At step 1, take two points at opposite ends of a diameter and label them both 1. At step n>1, place a point at the midpoint of each arc created at step n-1 and label it with the sum of the labels at the two adjacent points. What is the total sum of the labels after step n?
For example, after step 4 we have: 1, 4, 3, 5, 2, 5, 3, 4, 1, 4, 3, 5, 2, 5, 3, 4.
Solution
Answer: 2.3n-1.
True for n=1. The new points added at step n+1 have twice the sum of the points after step n, because each old point contributes to two new points. hence the total after step n+1 is three times the total after step n.
Problem 14
Given an isosceles triangle, find the locus of the point P inside the triangle such that the distance from P to the base equals the geometric mean of the distances to the sides.
Solution
Let the triangle be ABC, with AB=AC. Take the circle through B and C which has AB and AC as tangents. The required locus is the arc BC.
Suppose P lies on the arc. Let the perpendiculars from P meet BC in L, AB in N and AC in M. Join PB and PC. The triangles PNB and PLC are similar (PNB and PLC are both 90, and NBP = LCP because NB is tangent to the circle). Hence PN/PL = PB/PC. Similarly, triangles PMC and PLB are similar and hence PM/PL = PC/PB. Multiplying gives the required result PL2 = PM.PN.
If P is inside the circle and not on it, take P' as the intersection of the line AP and the arc. We have PL<P'L, but PM>P'M and PN>P'N, hence PL2<PM.PN. Similarly, if P is outside the circle and not on it, then PL2>PM.PN.
Solutions
Problem 1
Given 5 circles. Every 4 have a common point. Prove that there is a point common to all 5.
Solution
Let the circles be a, b, c, d, e. Let A be a point common to b, c, d, e, let B be a point common to a, c, d, e and so on. If any two of A, B, C, D, E coincide then the coincident point is on all 5 circles. Suppose they are all distinct. Then A, B, C are on d and e. Hence d and e coincide (3 points determine a circle). Hence D is on all 5 circles.
Problem 2
8 players compete in a tournament. Everyone plays everyone else just once. The winner of a game gets 1, the loser 0, or each gets 1/2 if the game is drawn. The final result is that everyone gets a different score and the player placing second gets the same as the total of the four bottom players. What was the result of the game between the player placing third and the player placing seventh?
Solution
The bottom 4 played 6 games amongst themselves, so their scores must total at least 6. Hence the number 2 player scored at least 6. The maximum score possible is 7, so if the number 2 player scored more than 6, then he must have scored 6 1/2 and the top player 7. But then the top player must have won all his games, and hence the number 2 player lost at least one game and could not have scored 6 1/2. Hence the number 2 player scored exactly 6, and the bottom 4 players lost all their games with the top 4 players. In particular, the number 3 player won against the number 7 player.
Problem 3
(a) The two diagonals of a quadrilateral each divide it into two parts of equal area. Prove it is a parallelogram.
(b) The three main diagonals of a hexagon each divide it into two parts of equal area. Prove they have a common point. [If ABCDEF is a hexagon, then the main diagonals are AD, BE and CF.]
Solution
(a) Let the quadrilateral be ABCD and let the diagonals AC, BD meet at E. Then area ABC = AC.EB.sin CEB/2, and area ADC = AC.ED.sin CEB/2, so E is the midpoint of BD. Similarly, it is the midpoint of AC. Hence the triangles AEB and CED are congruent, so angle CDE = angle ABE, and hence AB is parallel to CD. Similarly, AD is parallel to BC.
(b) Let the hexagon be ABCDEF. Let BE, CF meet at J, let AD, CF meet at K, and let AD, BE meet at L. Let AK=a, BJ=b, CJ=c, DL=d, EL=e, FK=f. Also let KL=x, JL=y and JK=z. Consider the pair of diagonals AD, BE. They divide the hexagon into 4 parts: the triangles ALB and DLE, and the quadrilaterals AFEL and BCDL. Since area ALB + area AFEL = area DLE + area BCDL, and area ALB + area BCDL = area DLE + area AFEL, the two triangles must have the same area (add the two inequalities). But area ALB = 1/2 AL.BL.sin ALB, and area DLE = 1/2 DL.EL.sin DLE = 1/2 DL.EL.sin ALB, so AL.BL = DL.EL or de = (a+x)(b+y). Similarly, considering the other two paris of diagonals, we get bc = (e+y)(f+z) and af = (c+z)(d+x). Multiplying the three inequalities gives: abcdef = (a+f)(b+y)(c+z)(d+x)(e+y)(f+z). But x, y, z are non-negative, so they must be zero and hence the three diagonals pass through a common point.
Problem 4
The natural numbers m and n are relatively prime. Prove that the greatest common divisor of m+n and m2+n2 is either 1 or 2.Solution
If d divides m+n and m2+n2, then it also divides (m+n)2 - (m2+n2) = 2mn and hence also 2m(m+n) - 2mn = 2m2 and 2n(m+n) - 2mn = 2n2. But m and n are relatively prime, so m2 and n2 are also. Hence d must divide 2.
Problem 5
Given a circle c and two fixed points A, B on it. M is another point on c, and K is the midpoint of BM. P is the foot of the perpendicular from K to AM.
(a) prove that KP passes through a fixed point (as M varies);
(b) find the locus of P.
Solution
(a) Take Y on the circle so that angle ABY=90. Then AY is a diameter and so angle AMY=90. Take X as the midpoint of BY. Then triangles BXK and BYM are similar, so XK is parallel to YM. Hence XK is perpendicular to AM, and so P is the intersection of XK and AM. In other words, KP always passes through X.
(b) P must lie on the circle diameter AX, and indeed all such points can be obtained (given a point P on the circle, take M as the intersection of AP and the original circle). So the locus of P is the circle diameter AX.
Problem 6
Find the smallest value x such that, given any point inside an equilateral triangle of side 1, we can always choose two points on the sides of the triangle, collinear with the given point and a distance x apart.
Solution
Answer: 2/3.
Let O be the center of ABC. Let AO meet BC at D, let BO meet CA at E, and let CO meet AB at F. Given any point X inside ABC, it lies in one of the quadrilaterals AEOF, CDOE, BFOD. Without loss of generality, it lies in AEOF. Take the line through X parallel to BC. It meets AB in P and AC in Q. Then PQ is shorter than the parallel line MON with M on AB and N on AC, which has length 2/3. If we twist the segment PXQ so that it continues to pass through X, and P remains on AB and Q on AC, then its length will change continuously. Eventually, one end will reach a vertex, whilst the other will be on the opposite side and hence the length of the segment will be at least that of an altitude, which is greater than 2/3. So at some intermediate position its length will be 2/3.
To show that no value smaller than 2/3 is possible, it is sufficient to show that any segment POQ with P and Q on the sides of the triangle has length at least 2/3. Take P on MB and Q on AN with P, O, Q collinear. Then PQ cos POM = MN - QN cospi/3 + PM cos pi/3. But PM>QN (using the sine rule, PM = OM sinPOM/sinOPM and QN = ON sinQON/sinOQN, but OM=ON, angle POM = angle QON, and angle OQN = angle OPM + pi/3 > angle OPM), and hence PQ > MN sec POM > MN.
Problem 7
(a) A 6 x 6 board is tiled with 2 x 1 dominos. Prove that we can always divide the board into two rectangles each of which is tiled separately (with no domino crossing the dividing line).
#(b) Is this true for an 8 x 8 board?
Solution
(a) We say a domino bridges two columns if half the domino is in each column. We show that for 0<n<6 the number of dominoes bridging columns n and n+1 must be at least 2 and even.
Consider first n=1. There cannot be 3 dominoes entirely in column 1, or it would be separately tiled. So there must be at least one domino bridging columns 1 and 2. The number must be even, because it must equal the number of squares in column 1 (even) less twice the number of dominoes (entirely) in column 1.
Now suppose it is true for n<5 and consider column n+1. There must be at least one domino bridging columns n+1 and n+2, or columns 1 thru n+1 would be separately tiled. The number must be even, because it must equal the number of squares in column n+1 (even) less the number bridging n and n+1 (even) less twice the number entirely in the column.
So in total there are at least 5 x 2 = 10 dominoes bridging columns. By the same argument there are at least another 10 bridging rows, but there are only 18 dominoes in total.
(b) No. For example:
1 2 3 3 1 1 2 2
1 2 1 2 2 3 3 1
3 3 1 3 1 2 4 1
1 2 2 3 1 2 4 3
1 3 3 2 2 1 2 3
3 2 1 1 4 1 2 1
3 2 3 2 4 3 3 1
1 1 3 2 1 1 2 2
Problem 8
Given a set of n different positive reals {a1, a2, ... , an}. Take all possible non-empty subsets and form their sums. Prove we get at least n(n+1)/2 different sums.
Solution
Assume a1 < a2 < ... < an. We have the following collection of increasing sums:
| a1 < a2 < ... < an | n sums | |
| a1+an < a2+an < ... < an-1+an | n-1 sums | |
| a1+an-1+an < a2+an-1+an < ... < an-2+an-1+an | n-2 sums | |
| ... | ||
| a1 + a2 + ... + an | 1 sum |
A total of 1+2+ ... +n = n(n+1)/2.
Problem 9
Given a triangle ABC. Let the line through C parallel to the angle bisector of B meet the angle bisector of A at D, and let the line through C parallel to the angle bisector of A meet the angle bisector of B at E. Prove that if DE is parallel to AB, then CA=CB.
Solution
The idea is to find an expression for the perpendicular distance h from D to AB. Let γ = ∠ACB, α = ½∠CAB, and β = ½∠ABC. We have h = AP sin α.
Using the sine rule on APC, we have AP = AC sin(γ+β)/sin(α+β), so h = AC sin α sin(γ+β)/sin(α+β). Similarly, the perpendicular distance k from E to AB is BC sin β sin(γ+α)/sin(α+β).
We also have that AC/BC = sin 2β/sin 2α, and hence h/k = sin 2β sin α sin(γ+β)/(sin 2α sin β sin(γ+α)). Using the fact that sin(γ+β) = sin(2α+β), and the expression for sin 2θ, we get h/k = (sin(2α+2β) + sin 2α)/(sin(2α+2β) + sin 2β) and hence h = k iff the triangle is isosceles.
For some reason the geometric solution took me longer to find. Let ED meet BC at X. Then XCD and XBE are isosceles, so BC = BX + XC = DX + XE = DE. Similarly, AC = DE. Hence AC = BC.
Problem 10
An infinite arithmetic progression contains a square. Prove it contains infinitely many squares.
Solution
Let the square be a2 and the difference d, so that all numbers of the form a2+nd belong to the arithmetic progression (for n a natural number). Take n to be 2ar + dr2, then a2+nd = (a+dr)2.
Problem 11
Can we label each vertex of a 45-gon with one of the digits 0, 1, ... , 9 so that for each pair of distinct digits i, j one of the 45 sides has vertices labeled i, j?
Solution
10 x 5 > 45, so some digit i0 must appear less than 5 times. But each occurrence can give at most 2 edges i0, j, so there are at most 8 edges i0, j, which is one too few.
Problem 12
Find all real p, q, a, b such that we have (2x-1)20 - (ax+b)20 = (x2+px+q)10 for all x.
Solution
Comparing coefficients of x20, we must have a = (220 - 1)1/20 (note that we allow either the positive or the negative root).
Set x=1/2. Then we must have (ax + b)20 = 0 = (x2+px+q)10, and hence ax + b =0 and x2+px+q = 0. So b = -a/2, and 1/4 + p/2 + q = 0.
Set x=0. Then we get q10 = 1 - b20 = 1/220, so q = 1/4 or -1/4, and p= -1 or 0 respectively. Comparing the coefficients of x19, we must have p = -1 and q = 1/4. So, if there is a solution, then it must be: a = (220 - 1)1/20, b = -a/2, p = -1, q = 1/4. This is indeed a solution because with these values, the lhs = 220(x - 1/2)20 - (x - 1/2)20a20 = (x - 1/2)20 = (x2 - x + 1/4)10 = rhs.
Problem 13
We place labeled points on a circle as follows. At step 1, take two points at opposite ends of a diameter and label them both 1. At step n>1, place a point at the midpoint of each arc created at step n-1 and label it with the sum of the labels at the two adjacent points. What is the total sum of the labels after step n?
For example, after step 4 we have: 1, 4, 3, 5, 2, 5, 3, 4, 1, 4, 3, 5, 2, 5, 3, 4.
Solution
Answer: 2.3n-1.
True for n=1. The new points added at step n+1 have twice the sum of the points after step n, because each old point contributes to two new points. hence the total after step n+1 is three times the total after step n.
Problem 14
Given an isosceles triangle, find the locus of the point P inside the triangle such that the distance from P to the base equals the geometric mean of the distances to the sides.
Solution
Let the triangle be ABC, with AB=AC. Take the circle through B and C which has AB and AC as tangents. The required locus is the arc BC.
Suppose P lies on the arc. Let the perpendiculars from P meet BC in L, AB in N and AC in M. Join PB and PC. The triangles PNB and PLC are similar (PNB and PLC are both 90, and NBP = LCP because NB is tangent to the circle). Hence PN/PL = PB/PC. Similarly, triangles PMC and PLB are similar and hence PM/PL = PC/PB. Multiplying gives the required result PL2 = PM.PN.
If P is inside the circle and not on it, take P' as the intersection of the line AP and the arc. We have PL<P'L, but PM>P'M and PN>P'N, hence PL2<PM.PN. Similarly, if P is outside the circle and not on it, then PL2>PM.PN.
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