Answers
There are many possible positions of @ahmadmangazap. However, for a participant to be considered as winner, they must have submitted at least one correct posiiton of @ahmadmangazap.
The easiest to get is (-4,10). Nevertheless, any participant who submitted any possible position of @ahmadmangazap in the original problem shall be considered as winner.
Solutions
Graphing method
We can visually obtain the position of @ahmadmangazap is to draw the points, and draw circles around @holovision with radius starting from 1 and increasing by 1, and @jfang003 with starting radius 2 and increasing by 2, until the circumferences of the circles touch each other at least once above 6 in the y-axis.
(12, 6.032) is a possible position of @ahmadmangazap based on the graph, though you have to use a big and precise graphing area (graphing apps such as Desmos is recommended).
Distance between A and H
= sqrt((10 - 12)^2 + (2 - 6.032)^2)
= sqrt((-2)^2 + (-4.032)^2)
= 4.5008
Distance between A and J
= sqrt((3 - 12)^2 + (6 - 6.032)^2)
= sqrt((-9)^2 + (0.032)^2)
= 9.0001
Because we used approximation in 6.032, we get rounding error in the distances. However, that is an acceptable error.
Midpoint method
Just place @ahmadmangazap somewhere on the graph such that @jfang003 becomes the midpoint of @ahmadmangazap and @holovision.
The midpoint formula is (xm, ym) = ((xa + xb)/2, (ya + yb)/2), such that:
- (xm,ym) = @jfang003's position which is (3,6)
- (xa, ya) = @ahmadmangazap's position which is unknown
- (xb, yb) = @holovision's position which is (10,2)
Our equation is now (3,6) = ((xa + 10)/2, (ya + 2)/2). In the equation, xa = -4 and ya = 10. Therefore, a possible position of @ahmadmangazap is (-4,10).
Systemic method
Nevermind the solution below, since I believe I already explained enough using the easier solutions above, unless you want 🤯
This method involves systems of equations.
- Let's use capital letters as positions of the persons. A for @ahmadmangazap, H for @holovision, and J for @jfang003.
- Let (x,y) be the coordinates of @ahmadmangazap.
Let d be the distance of @ahmadmangazap from @holovision
Let 2d be the distance of @ahmadmangazap from @jfang003
We use the distance formula to get the distance between any two points. The distance formula is
the distance is equal to the square root of the sum of the squares of the difference between the change in y and change in x
Using the distance formula and the two distance equations above, we get the two equations:
- d^2 = (10 - x)^2 + (2 - y)^2
- d^2 = x^2 - 20x + 100 + y^2 - 4y + 4
- Equation 1: d^2 = x^2 - 20x + y^2 - 4y + 104
- (2d)^2 = (3 - x)^2 + (6 - y)^2
- (2d)^2 = x^2 - 6x + 9 + y^2 - 12y + 36
- Equation 2: 4d^2 = x^2 - 6x + y^2 - 12y + 45
Since the d on Equation 1 is the same as d on Equation 2, we can substitute Equation 1 to Equation 2.
- x^2 - 6x + y^2 - 12y + 45 = 4 * (x^2 - 20x + y^2 - 4y + 104)
- x^2 - 6x + y^2 - 12y + 45 = 4x^2 - 80x + 4y^2 - 16y + 416
- 3x^2 - 74x + 3y^2 - 4y + 371 = 0
- x^2 - 74x/3 + y^2 - 4y/3 + 371/3 = 0
- (x^2 - 74x/3 + 1369/9) + (y^2 - 4y/3 + 4/9) + 371/3 - (1369/9 + 4/9) = 0
- (x + 37/3)^2 + (y - 2/3)^2 - 260/3 = 0
- Equation 3: (x + 37/3)^2 + (y - 2/3)^2 = sqrt(260/3)^2
We can use any pair of x and y values satisfying Equation 3 for as long as y is greater than 6. The higher of @jfang003 and @holovision's positions has a y-intercept of 6.
Winner: none 🤯
1 HIVE shall be distributed to the previous winners in proportion to the number of days they correctly answered first.
This explanation took me about 3 hours to finish! 🤯🤯
Mentions: @ahmadmanga (@ahmadmangazap), @jfang003, @holovision, @jwynqc, and
@appukuttan66
Special mentions: @dbuzz, @chrisrice, @jancharlest 🤯