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Since the angles are the same, there are 2 similar triangles: \(\triangle CPB\sim\triangle DMB\). This means that \[{CB\over DB}={PB\over MB}\Rightarrow DB\cdot PB=CB\cdot MB\] Now we can compute some triangle areas using the formula from side-angle-side. \[\text{area}(PDB) = {1\over2}\cdot DB\cdot PB\cdot\sin(B)\] \[\text{area}(ACB) = {1\over2}\cdot CB\cdot AB\cdot\sin(B) =CB\cdot MB\cdot\sin(B)\] Here we have used \(2\cdot MB=AB\). By comparing these formulas, we see that the area of \(\triangle ACB\) is twice the area of \(\triangle PDB\). The area of \(ACDP\) is just the area of \(\triangle ACB\) minus area of \(\triangle PDB\). Therefore, they have the same area so the ratio is 1.
How many \(\triangle ABC\) can you draw so \(\overline{AB}=5\) units, \(\overline{AC}=2\) units, and \(\text{area}(ABC)=4\) units2
Using the side-angle-side area formula, we have \(4={1\over2}\cdot5\cdot2\cdot\sin(A)\). This equation tells us \(\sin(A)={4\over5}\) which has 2 solutions for \(0^\circ<A<180^\circ\) which are about \(53^\circ,127^\circ\).
\(y^2=x^3-2\) with \(x,y\in\mathbb{Z}\)
By guessing a few numbers, we can see that \(x=3\) with \(y=\pm5\) are solutions. But to show these are the only solutions, we need some more advanced theory. This Diophantine equation is a case of Mordell's equation. See theorem 3.4 in this paper for a solution to this exact problem. It uses properties of the group \(\mathbb{Z}[\sqrt{-2}]=\{a+b\sqrt{-2}:a,b\in\mathbb{Z}\}\).
Xander jogged to and from the park at 5mph. The next day, he jogged at 4mph to the same location and 6mph back. The second job took 2 minutes longer than the first. How many miles did Xander jog the first day?
Let \(d\) be the distance between the park and where he started. On day 1, he spent time \(t_1=2d/5mph\). On day 2, we can write an equation for his time: \[t_2={d\over4mph}+{d\over6mph}={10d\over24mph}=t_1+2min={2d\over5mph}+2min\] Now let's switch all the units to hours and miles. The equation is \[{10d\over24}={2d\over5}+{1\over30}\Rightarrow d=2\] This tells us that half of the distance is 2 miles so the total distance is 4 miles.
Find the largest prime number whose base 4 representation only has the digit 1.
In base 4, all numbers with only the digit 1 are of the form \(1+4+4^2+\ldots+4^n={4^{n+1}-1\over4-1}={1\over3}(4^{n+1}-1)\). If \(n=1\), then we have 5 which is prime. Now we show that if \(n>1\), the number is not prime. We can factor \[{1\over3}(4^{n+1}-1)={1\over3}(2^{n+1}+1)(2^{n+1}-1)\] where we will have \(2^{n+1}+1\geq9\) and \(2^{n+1}-1\geq7\). Both of these differ by 2 so 3 can divide only 1 of them, leaving a nontrivial factorization. Therefore, the answer is 5.
\(10!\) seconds is exactly how many weeks?
First we start with the factorization of \(10!=2^8\cdot3^4\cdot5^2\cdot7\). Now we can find how many seconds are in a week. \(60\cdot60\cdot24\cdot7 =2^7\cdot3^3\cdot5^2\cdot7\). Then dividing, we find it is \(2\cdot3=6\) weeks.
\(x=z^2-y^3+5w^2+9z\) where \(w=2,y=5,z=7\).
This is simply substituting some numbers. \[x=7^2-5^3+5\cdot2^2+9\cdot7=49-125+20+63=-76+83=7\]
What is the largest cube in the Virahanka-Fibonacci sequence?
By looking at the first few terms with \(F_0=0,F_1=1,F_n=F_{n-1}+F_{n-2}\), we will see the values \(1,8,144\). Here, 8 is a cube. Showing that there are no more is a bit challenging. There is a 2006 paper by Andrejic that discusses the conjecture, available here. In 2006, a proof was published, available here.
\(123456789\mod45\)
We can compute this manually, but there are some ways to make it easier. One thing to notice is that \(10^n\equiv10\mod45\) when \(n\geq2\). Therefore, \(123456789\equiv9+10(1+2+3+4+5+6+7+8)=9+8\cdot45\equiv9\mod45\). We could also observe that \(123456789\equiv0\mod9\) and \(123456789\equiv4\mod5\) and find the same result using the Chinese remainder theorem.
\(1594593=y^y(y^x+x)\) with \(x,y\in\mathbb{Z}\)
We can start with the prime factorization \(1594593=3^3\cdot7\cdot11\cdot13\cdot59\). By looking at this, we can only match the \(y^y\) part with \(y=1\) or \(y=3\). If \(y=1\), then we have a solution with \(x=1594592\) which is not the intended solution. If \(y=3\), we can see that the remaining part is \(7\cdot11\cdot13\cdot59=59059=3^{10}+10\) so the intended solution has \(x=10\).
Let \(n_i\) be the binary number with \(i\) 1's (and no 0's). Find the sum of the binary digits of \(n_{44}/n_4\).
A binary number with \(i\) 1's is \(2^i-1\). Therefore, we are looking for the sum of binary digits of \[{2^{44}-1\over2^4-1}={(2^4)^{10}-1\over2^4-1}=1+2^4+2^8+\ldots+2^{40}\] This is 11 terms for the binary digits so the solution is 11.
The number of non-negative integer solutions to \(\sqrt{a}+\sqrt{b}=\sqrt{4477}\).
Square both sides and we have \(a+b+2\sqrt{ab}=4477=11^2\cdot37\). Since \(\sqrt{ab}\) cannot be a .5 number, it must be an integer. We can factor \(a\) and \(b\) each into some perfect square and a part common to each since a product of distinct primes dividing one must divide the other. Therefore, we have \(a=x^2y,b=yz^2\) for some \(x,y,z\). Substituting into the original equation, we get \(x\sqrt{y}+z\sqrt{y}=11\sqrt{37}\) or \((x+z)\sqrt{y}=11\sqrt{37}\). The integer solutions will come from \(x+z=11\) so there are 12 solutions we can choose in nonnegative integers.
\(23_{10}\cdot41_{10}=577_x\)
Writing an equation for this, we have \(23\cdot41=5x^2+7x+7\) and we need a base \(x\geq8\). This is \(943=5x^2+7x+7\Rightarrow5x^2+7x-936=0 \Rightarrow(5x+72)(x-13)=0\Rightarrow x=-72/5,13\). Therefore, the solution is base 13.
A figurine requiries 1 ounce of paint to decorate. How many ounces of paint are needed to paint 350 similar figurines that are \(1/5\) the height of the original?
By scaling down a factor of 5, the surface areas scale down by a factor of 25. Therefore, 25 of the smaller figures require 1 ounce. Since \(350=14\cdot25\), the answer is 14.
How many four digit numbers that are divisible by 9 only use the digits \(0,1,2,3\)?
This problem has a problem. The answer is not 15 and it will be explained.
We can consider this by cases and counting. If the numbers start with 3, then we can form the remaining digits to sum to 6 by using 330, 321, 222. With permutation counting, this is 3+6+1=10 such numbers. If the first digit is 2, the remaining can be from 331 or 322, so 3+3=6 numbers. If the first digit is 1, then we need digits 332 so there are 3 more. In total, there are 19 numbers.
This can be confirmed with the following python code.
>>> [x for x in range(1000,10000) if set(str(x)) <= set('0123') and sum(map(int,str(x)))==9]
[1233, 1323, 1332, 2133, 2223, 2232, 2313, 2322, 2331, 3033, 3123, 3132, 3213, 3222, 3231, 3303, 3312, 3321, 3330]
The volume enclosed by the paraboloids \(z=w^2+y^2\) and \(z=8-w^2-y^2\) is \(\pi x\).
In cylindrical coordinates, we have \(r^2=w^2+y^2\). So the parabolas intersect when \(r^2=4\Rightarrow r=2\). We can integrate the difference on the disk \(r\leq2\). \[\begin{align}&\int_0^{2\pi}\int_0^2(8-r^2-r^2)rdrd\theta =2\pi\int_0^2(8r-2r^3)dr=2\pi\left[4r^2-{r^4\over2}\right]_0^2\\ &=2\pi\left[16-{16\over2}\right]=16\pi\end{align}\] Therefore \(x=16\).
\(f(z)=z^7-16z^6+z^2+8z-1\), \(w=\text{the sum of the roots of}\ f(z)\), \(y=\text{the product of the roots of}\ f(z)\), \(x=w+y\).
Using Vieta's formulas, the sum of roots is \(w={-a_{n-1}\over a_n} =-{-16\over1}=16\) and the product of roots is \(y=(-1)^n{a_0\over a_n} =-{-1\over1}=1\). Therefore, \(x=17\).
How much would you save if you bought a widget with a $150 list price that is 12% off?
You would save 12% of the price which is \(0.12\cdot150=18\).
\(f:\mathbb{R}\to\mathbb{R}\) satisfies for all \(s,t\in\mathbb{R}\) \[f(s^3)+f(t^3)=(s+t)\cdot(f(s^2)+f(t^2)-f(st))\] Find \[\quad x={f(100)\over f(5)}\]
By using \(s=t=0\), we find \(f(0)=0\). By using \(s=-t\), we find \(f(s^3)+f(-s^3)=0\) which cells us that \(f\) is odd because \(s^3\) goes over all of \(\mathbb{R}\).
A questionable idea is to do the following. Substitute \(s=a^{1/3}\). Then we get \(f(a)=a^{1/3}f(a^{2/3})\). By repeatedly substituting, we get: \[f(a)=a^{1/3}f(a^{2/3})=a^{1/3+(1/3)(2/3)}f(a^{(2/3)^2}) =a^{1/3+(1/3)(2/3)+(1/3)(2/3)^2}f(a^{(2/3)^3})=\ldots\] So as we extend this, we have \[f(a)=a^{{1\over3}\left(1+{2\over3}+\left({2\over3}\right)^2 +\ldots+\left({2\over3}\right)^n\right)}f(a^{(2/3)^n})\] As \(n\to\infty\), we can evaluate this to be \(f(a)=a\cdot f(1)\), that is, \(f\) is a linear function with \(f(0)=0\). Then using this property for \(x\), \[x={f(100)\over f(5)}={100f(1)\over5f(1)}={100\over5}=20\] Here, of course, \(f(1)\neq0\). If \(f(1)=0\) then \(f(5)=0\) so it would not make sense so this assumption is reasonable.
This is a kite so the area is half of the product of the lengths connecting opposite corners. The area is \({1\over2}\cdot6\cdot7=21\).
How many ways can you partition 8?
This can be solved by recursion or dynamic programming. We can define a recurrence \(P(n,i)\) as the number of ways to partition \(n\) using positive integers \(\leq i\). This can be broken up into 2 cases. If the largest number used in a partition is \(i\), then use \(P(n-i,i)\). If the largest number is smaller than \(i\), then use \(P(n,i-1)\). This tells us the recurrence is \(P(n,i)=P(n-i,i)+P(n,i-1)\). The base cases are \(P(n,1)=1\) and \(P(1,i)=1\). We can fill in the following table going row by row using the recurrence.
| \(n\textbackslash i\) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 2 | 2 | 2 | 2 | 2 | 2 |
| 3 | 1 | 2 | 3 | 3 | 3 | 3 | 3 | 3 |
| 4 | 1 | 3 | 4 | 5 | 5 | 5 | 5 | 5 |
| 5 | 1 | 3 | 5 | 6 | 7 | 7 | 7 | 7 |
| 6 | 1 | 4 | 7 | 9 | 10 | 11 | 11 | 11 |
| 7 | 1 | 4 | 8 | 11 | 13 | 14 | 15 | 15 |
| 8 | 1 | 5 | 10 | 15 | 18 | 20 | 21 | 22 |
So the value we are looking for is \(P(8,8)=22\). Alternatively we can enumerate them which is not much more work for this problem. There is no known closed form solution to counting integer partitions.
Find the smallest \(x\) such that \(49\) divides \(1^2+2^2+\ldots+x^2\).
The sum of squares formula is \[1^2+2^2+\ldots+x^2={x(x+1)(2x+1)\over6}\] We will need the factor 7 to appear twice in the numerator. If \(x=24\), then \(2x+1=49\) so the solution is at most 24. If \(7\mid x\) then \(7\nmid(x+1)\) and \(7\nmid(2x+1)\) so we would need \(49\mid x\) for a solution. If \(7\mid(x+1)\) then \(7\nmid x\) and \(7\nmid(2x+1)\) so we would need \(49\mid(x+1)\). Both of these would give bigger solutions. By similar reasoning we can also show that \(7\mid(2x+1)\) implies \(7\nmid x\) and \(7\nmid(x+1)\). Therefore the solution is \(x=24\).
Using the Pythagorean theorem, \(x=\sqrt{65^2-60^2}=25\). We can also observe that this right triangle is scaled up from the 5-12-13 right triangle by a factor of 5.
For a regular hexagon \(S\), find the number of equilateral triangles that have at least two vertices that are also vertices of \(S\).
First, there are 12 triangles when we select 2 adjacent hexagon vertices, shown below, 6 inside, and 6 outside.
If we select 2 vertices with 1 between, we can form 2 inner triangles shown below. There are 6 ways to choose 2 vertices such that there is 1 between them, but some choices give the same triangle.
Using the same vertex choices (2 with 1 between), there are 6 outer triangles, shown below with 1 example highlighted in green.
Finally, we can pick opposite vertices and form 6 more triangles, shown below with an example highlighted in green.
In total, we count \(12+2+6+6=26\) triangles.
Find the sum of the coefficients of \((1+x+x^2)^3\).
This can be done with the multinomial theorem. Since all the coefficients of \(x\) powers are 1, we only have to care about the multinomial coefficients. We can form sums to 3 in the following ways (ignoring order): \[3+0+0,\quad2+1+0,\quad1+1+1\] Summing the multinomial coefficients multiplied by how many ways we can reorder these sums, we have \[{3\choose1,2}{3\choose3,0,0}+{3\choose1,1,1}{3\choose2,1,0} +{3\choose3}{3\choose1,1,1}=3\cdot1+6\cdot3+1\cdot6=3+18+6=27\]
One way to view this more generally is considering the sum of coefficienst of \((1+x+x^2+\ldots+x^{n-1})^n\). If we expand this, we get \(n^n\) terms, each a power of \(x\) with coefficient \(1\). Summing them, the sum of coefficients would just be \(n^n\). For this problem, \(n=3\).
Find 108 times the probability that 3 sequential die rolls are in non-decreasing order.
Out of the \(6^3=216\) possible 3 die rolls, each nondecreasing order of 3 rolls corresponds to a way of selecting 3 numbers from 1-6 with replacement, where order does not matter (since we need them to be sorted nondecreasing and repetitions are allowed). There are \({6+3-1\choose3}={8!\over3!5!} ={8\cdot7\cdot6\over6}=56\) such ways. Therefore, the probability of the rolls being in nondecreasing order is \(56/216=28/108\) and multiplying this by 108 gets the answer 28.
Simplify to \(37\equiv8\mod x\). Then \(x\mid(37-8)\Rightarrow x\mid29\). Since 29 is prime, \(x=1\) or \(x=29\). The writers intended for the solution be be \(x=29\).
Find the number of two card hands where both cards are red and have the rank king, queen, or jack.
There are 3 ranks and 2 suits, so 6 possible cards. Then we choose 2, but \[{6\choose2}={6\cdot5\over2}={30\over2}=15\] If order matters (which it has to for us to get the expected answer), then it is permutations of 2 of the cards so \(6\cdot5=30\).