Sobhan's Toothpick Sum

We have two continuous variables determining if our event occurs: the length of the longer piece of the first toothpick (call it $x$ ) and the length of the longer piece of the second toothpick (call it $y$ ). This means we can't use counting techniques to find the probability, but we can use area ratios.

We know that the possible values for the lengths of the longer pieces of the toothpicks are elements in $\left[.5,1\right]$ . Without loss of generality (because the two toothpicks are identical), let $y>x$ . Our sample space then is the region bounded below by $y=x$ , above by $y=1$ , and to the left by $x=.5$ . We need to find the area in which our event (that $x,y$ are both greater than the sum of the lengths of the smaller pieces, $1-x+1-y=2-x-y$ ) occurs, then divide this area by the total area of our sample space.

The intersection of the portion of the plane satisfying this inequality and our sample space is bounded below by $y=x$ and $y=2-2x$ and above by $y=1$ . The intersection of $y=2-2x$ and $y=x$ occurs at the point $\frac{2}{3},\frac{2}{3}$ .

Our desired area ratio is equal to the ratio of the lengths of the segments from $(\frac{2}{3},\frac{2}{3})$ to $(1,1$ ) and from $(.5,.5)$ to $(1,1)$ because both the event triangle and the sample space triangle share the altitude from $(.5,1)$ to the line $y=x$ . The ratio of the lengths is $\frac{\sqrt{2}}{3}/\frac{\sqrt{2}}{2}=\frac{2}{3}$ , leaving a final answer of $2+3=\boxed{5}$ .

I supposed that the toothpicks were of length 1. With this, I change the problem, looking for the area of the set $\:A\subset[0,1]\times [0,1]$ of points $(p,q)$ such that

$\min\{p,1-p\}+\min\{q,1-q\}<\max\{p,1-p\}$ , and $\min\{p,1-p\}+\min\{q,1-q\}<\max\{q,1-q\}$ .

Since $\min\{x,1-x\}$ has two options, we have four cases. I’m going to solve one case and the other three are symmetric.

Case 1. $\:p=\min\{p,1-p\}$ and $\:q=\min\{q,1-q\}$ , then i need to find the region of points such that $\:p+q<1-p,$ and $\:p+q<1-q.$ This is equivalent to the set $\{(p,q) : \:q<1-2p,\;q<\frac{1}{2}-\frac{p}{2}\}\cap([0,1]\times[0,1]).$

This set has area is $\frac{1}{6}$ . Since we have four sets, the area of the set $A$ is equal to $4\times\frac{1}{6}=\frac{2}{3}$ , therefore the answer is $2+3=5$ .

Let the parts of the first toothpick have lengths of a and b, and parts of the second toothpick have lengths c and d. Obviously, a + b = c + d. Without loss of generality, let a \leq b, c \leq d, and a + b = c + d=1. To satisfy the condition above, a + c \leq d \Rightarrow a + 2c \leq 1. Also, a + c \leq b \Rightarrow 2a+c \leq 1.

Draw a graph with y-axis as a and x-axis as c. The two equations that now bond the area of probability is x + 2y \leq 1 and 2x + y \leq 1. Thus, by drawing out the area of probability, we find that it is a quadrilateral bonded by points (0,0), (0, \frac {1}{2}), (\frac {1}{3},\frac {1}{3}) and (\frac {1}{2},0) , which is of area \frac {1}{6}.

Without the condition that the two largest pieces are each larger than the sum of the lengths of the the two smallest pieces, the area of probability is only bonded by a \leq b, c \leq d, and a + b = c + d = 1 \Rightarrow a \leq \frac {1}{2} and c \leq \frac {1}{2}. Again, by drawing a graph with y-axis as a and x-axis as c, we find that the area of probability has area \frac {1}{4}.

Thus, the probability that the two largest pieces are each larger than the sum of the lengths of the the two smallest pieces is \frac {\frac {1}{6}}{\frac {1}{4}} = \frac {2}{3}, so the answer is 2 + 3 = 5

WLOG, let the toothpicks have length $1$ . After dropping, let the two toothpicks break into pieces of length $x$ and $1-x$ and $y$ and $1-y$ , respectively, with $x,y\ge\frac12$ . Because the pieces with lengths $x$ and $y$ are the two largest ones, we have $x\ge (1-x)+(1-y)\Rightarrow 2x+y\ge2$ and $y\ge (1-x)+(1-y)\Rightarrow x+2y\ge2$ . The region bounded by these inequalities and $\frac12\le x\le 1$ and $\frac12\le y\le1$ has area $\frac12\cdot\frac13=\frac16$ . The total possible area is $\frac12\cdot\frac12=\frac14$ , so our answer is $\dfrac{\frac16}{\frac14}=\frac23$ .

Let the lengths of the smaller pieces be x and y so $4\int_0^{1/2}\int_0^{1/2}1\,dy\,dx=1$

The condition that both larger pieces are larger than the sum of the smaller pieces is $x+y<1-x\wedge x+y<1-y$ or $y<1-2x\wedge y<\frac{1-x}{2}$ and so the chance that this holds is simply $p=4\int_0^{1/2}\int_0^{\min(1-2x,(1-x)/2)}1\,dy\,dx$ which can be split into parts at the point that $1-2x=\frac{1-x}{2}$ or 1/3, thus $p=4\int_0^{1/3}\int_0^{(1-x)/2}1\,dy\,dx+4\int_{1/3}^{1/2}\int_0^{1-2x}1\,dy\,dx$ which can be evaluated as usual: $\begin{array}{c}p=\int_0^{1/3}(2-2x)\,dx+\int_{1/3}^{1/2}(4-8x)\,dx\\ =\left.(2x-x^2)\right|_0^{1/3}+\left.(4x-4x^2)\right|_{1/3}^{1/2}\\ =(2\cdot1/3-1/3^2)-(2\cdot0-0^2)+(4\cdot1/2-4\cdot1/2^2)-(4\cdot1/3-4\cdot1/3^2)\\ =5/9-0+1-8/9=2/3 \end{array}$ and so p=2/3 and 2+3=5.

Suppose that the toothpick are of length 1, the first toothpick breaks into two pieces of length $x,1-x \, (0.5 \le x \le 1)$ , the secend toothpick breaks into two pieces of length $y,1-y \, (0.5 \le y \le 1)$ .

The two largest pieces are each larger than the sum of the lengths of the the two smallest pieces if and only if we have $x>1-x+1-y, y>1-x+1-y$ , or equivalently, $2x+y>1, x+2y>1$ . On the plane with cartesian coordinates $Oxy$ , we draw two half-planes $A : 2x+y>1, B : x+2y>1$ and two regions $C : 0.5 \le x \le 1, D : 0.5 \le y \le 1$ . The region $\mathcal{U}$ bounded by the two regions $C,D$ consists of all points $(x,y)$ with possible coordinates, the region $\mathcal{V}$ bounded by $A,B,C,D$ consists of all points $x,y$ with coordinates satisfy the given conditions.

We have $S_\mathcal{U}=1/4$ and $S_\mathcal{V}=1/6$ . Therefore, the probality is $4/6=2/3$ .

Assume that the length of each whole toothpick is 1. Let x and y be the lengths of the two shorter pieces. Then, we have:

$1 - x \ge x + y \rightarrow 1 >= 2x + y$ $1 - y \ge x + y \rightarrow 1 >= x + 2y$

Note that (x,y) lies within the square with opposite corners (0, 0) and (1/2, 1/2). Our valid (x, y) also lie under the lines 2x + y = 1 and x + 2y = 1, so we have to compute the area of a convex quadrilateral. There are multiple ways to do so, but an algebraic way to calculate uses Shoelace Theorem (since I don't know how to draw diagrams). The intersection of the two lines is:

2x + y = 1 = x + 2y x = y x = 1/3 = y

Coordinates: (0, 0) (0, 1/2) (1/3, 1/3) (1/2, 0)

Area = $1/2 * \left| ( (0 * 1/2) + (0 * 1/3) + (1/3 * 0) ) - ( (0 * 0) + (1/2 * 1/3) + (1/3 * 1/2) ) \right|$

Area = 1/2 * | 0 - (1/3) | = 1/6

Thus, the probability is (1/6) / (1/4) = 2/3, so the answer is 2 + 3 = 5.

Let a and b be the smaller parts that are broken off. We know 1-a>a+b and 1-b>a+b. This yields two conditions that must be satisfied: b<1-2a and b<1-a/2. Graphing these two inequalities and finding the area of the overlapping regions, you get an area of 1/6 which is 2/3 of the total area which is 1/4.

Define the length of a toothpick to be 1 unit. Let x and y be the lengths of the larger pieces; the smaller pieces are then (1 - x) and (1 - y). To satisfy the conditions of the scenario, x and y must each be larger than the sum of (1 - x) and (1 - y). Also, x and y cannot exceed 1.

The four inequalities are then: x < 1, y < 1, (1-y) + (1-x) < x, and (1-y) + (1-x) < y. Solving the latter two for x and y, respectively, we obtain: \frac {2-y}{2} < x and \frac {2-x}{2} < y.

The area within these boundaries (equivalently, the sum of the points (x,y) that satisfy our conditions) is \frac {1}{6}. Since the larger piece of each toothpick can occur on either side, we have (2 toothpicks) \times (2 sides per toothpick) \times \frac {1}{6} = \frac {2}{3} = \frac {a}{b}. a+b = 5.

We discuss two possible cases as following:

1.Both breaks occur at a position no more than one third from either end of the toothpick.

Then the probability of both of these happening is $P_1=\frac{2}{3} \times \frac{2}{3}=\frac{4}{9}$

2.One break occurs at the middle one third length of the toothpick.

In this case, the probability of the second break to satisfy the condition will vary with the change of position of the first break. Denote one of the one third point of the first toothpick as 0, and the other as $\frac{1}{3}$ . Denote the position of the first break as x.

We can assume that both breaks occur at either half of the toothpicks.

The probability of the second break to satisfy the condition depends on the ratio between the maximum acceptable length $l$ from the end and the total length of the toothpick. When $x \leq \frac{1}{6}$ , i.e. on either side of the middle of the toothpick, $l+2x=\frac{1}{3}$ . This gives $P=\int_0^\frac{1}{6} \mathrm{l}\,\mathrm{d}x=\int_0^\frac{1}{6} \mathrm{=-2x+\frac{1}{3}} \mathrm{d}x=\frac{1}{18}$

Furthermore, the second break may happen from both end of the second toothpick while the first break may happen from both side from the centre. This means the total probability for case 2 $P_2=4P=\frac{2}{9}$ .

In conclusion, the total probability that satisfies the condition is (P_{total}=\frac{2}{9}+\frac{4}{9}=\frac{2}{3}. Therefore, the answer is 2+3=5.

Let x and y scalar values denote the position of the breaking points for the first and second toothpicks and the pieces $[0$ $\mathrm x]$ and $[0$ $\mathrm y]$ be the short sides of their respective toothpicks (this is an easy assumption, since rotating the toothpick would change the placement of the short end).

Let us also assume without loss of generality that the length of the toothpicks be 2 units. The uniform probability distributions are easy to see $\mathrm x$ ~ $\cup[0$ $1]$ and $\mathrm y$ ~ $\cup[0$ $1]$ .

If $\mathrm x \leq \frac{2}{3}$ , the values that y can take is in the interval of $[0$ to $1-\frac{\mathrm x}{2}]$

If $\mathrm x > \frac{2}{3}$ , the values that y can take is in the interval of $[0$ to $2-2\mathrm x]$

We integrate the values of possible y over all the values of x to find the probability.

$\int_0^\frac{2}{3} \mathrm{(1-\frac{\mathrm x}{2})}\,\mathrm{d}x$ + $\int_\frac{2}{3}^1 \mathrm{(2-2\mathrm x)}\,\mathrm{d}x$

= $\left.(\mathrm x - \frac{\mathrm x^2}{4})\right|_0^\frac{2}{3} + \left.(2\mathrm x - \mathrm x^2)\right|_\frac{2}{3}^1$

= $\big(\frac{2}{3} - \frac{1}{9}\big) + \big((2-1) - (\frac{4}{3} - \frac{4}{9})\big)$

= $\frac{5}{9} + \frac{1}{9} = \frac{2}{3}$

Thus, $\mathrm a = 2$ and $\mathrm b = 3$ , and $\mathrm a + \mathrm b = 5$

We can represent the probability space as a unit square, where the $x$ -axis represents where the first toothpick breaks and the $y$ -axis represents where the second toothpick breaks. We let $x$ be the position of the first break and $y$ the position of the second break. There are four possibilities depending on whether each of $x$ and $y$ is larger or smaller than $\frac{1}{2}$ .

We will consider the case $x,y < \frac{1}{2}$ and the other cases will follow by symmetry. We observe that the two smallest pieces will have lengths $x,y$ , and the larger pieces will have lengths $1-x,1-y$ . In order for the sum of the lengths of the small pieces to be less than the length of the larger pieces, we must have $x + y < 1 - x$ and $x + y < 1 - y$ . This is equivalent to $2x + y < 1$ and $2y + x < 1$ .

In the figure below, we shade the region where both equations are satisfied, and symmetrically add in the other regions.

The two lines $2x + y = 1$ and $2y + x = 1$ intersect at the point $\left(\frac{1}{3},\frac{1}{3} \right)$ , so one shaded region consists of a square of side length $\frac{1}{3}$ and two right triangles with side lengths $\frac{1}{3}$ and $\frac{1}{6}$ . So the total area of this shaded region is $\frac{1}{3}\times \frac{1}{3} + \frac{1}{3}\times \frac{1}{6} = \frac{1}{6}$ . Then the total shaded area is $4 \times \frac{1}{6} = \frac{2}{3}$ . Thus, the value of $a + b$ is $2 + 3 = 5$ .

first we calculate where one of the toothpick should break so that the given condition is fulfilled no matter at which point the other toothpick breaks.

we get that if one of the toothpick breaks at 3/4 or more of its size the condition remains fulfilled no matter at which point the other toothpick breaks. the probability that one of the toothpick breaks at 3/4 of its size or more is 1/4.we also realize that if the first toothpick doesn't break at 3/4 of its size or more the second one will have to. Thus either of the two toothpicks will have to break at 3/4 of their size or more for the condition to be fulfilled and the probability of that happening is 1/4 which gives a+b=5

Suppose a and b are the larger parts and c is the sum of two shorter parts. The only expected measurements for this case are a > c a < c Or b > c b < c The probability picking a > c and b>c is 1/2 times 1/2 = 1/4 The sum is 5

Assume that both toothpicks break with equal part lengths (the length of the long ends $L_l$ are equal and the length of the short ends $L_s$ are equal). In the long run, this is is expected to happen.

The probability of each of the long ends being longer than the sum of the short ends can be denoted as $P\left(L_l>2L_s\right)$ and is related by the equation $P\left(L_l>2L_s\right) = \frac{L_l}{L_T}$ .

Since the length of the short ends are equal to the total length minus the length of the long ends, $L_l > 2L_s = 2\left(L_T-L_1\right) = 2L_T-2L_1$ . Using properties of inequalities, it can be seen that $L_T < \frac{3}{2}L_l$ .

By plugging these results into the formula, it can be seen that $P\left(L_l>2L_s\right) = \frac{L_l}{3L_l/2} = \frac{2}{3}$ . Summing the numerator and the denominator yields $2+3=5$ .