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The large left triangle is equilateral with side length \(x\) as given. By using angle properties and observing the upper small triangle is isosceles, we can show the other large triangle and the 2 small ones must also be equilateral. The other part we have to assume is that the 2 small triangles are congruent, which is not stated. It is possible to draw this diagram where the small triangles are different sizes but a side from each add to the same value which can be shown to be \(4/\sqrt{3}-2\).
Next, focus on the following 30-60-90 triangle (orange).
Its right side is \(\sqrt{3}-1\) since it is part of the square. The top side is \(1-1/\sqrt{3}\) and the hypotenuse is \(2-2/\sqrt{3}\). Let \(y\) be the side length of the small triangles which we have to assume are congruent. Then an equation for \(x\) is \[x=2-{2\over\sqrt{3}}+y\] Now along the top edge of the square, another equation for \(x\) we can form is \[x=1-{1\over\sqrt{3}}+\sqrt{3}-1-y\] The value of \(y\) is not important to find \(x\) because we can add the 2 equations to cancel it \[2x=2-{2\over\sqrt{3}}+1-{1\over\sqrt{3}}+\sqrt{3}-1 =3-{3\over\sqrt{3}}+\sqrt{3}-1=3-\sqrt{3}+\sqrt{3}-1=2\] Therefore, \(x=1\). If we want, \(y=2/\sqrt{3}-1\).
An alternative more complicated solution can be found by forming a system of equations with the following side lengths.
Let \(w=\sqrt{3}-1\) be the square side length. Then from the sides, we can find \(x+y=z+w\). The total are of the 2 large triangles can form the following equation \[{x^2\sqrt{3}\over4}=w^2+{y^2\sqrt{3}\over4}+zw\] And finally for the orange triangle, \((x-y)^2=z^2+w^2\). These equations are tough to solve. A computer algebra system found 4 solutions only one of which is valid because all the side lengths must be positive.
There is symmetry that can be observed with the substitutions \(u=2^x,v=5^x\) \[{u^3+v^3\over u^2v+uv^2}={(u+v)(u^2-uv+v^2)\over uv(u+v)} ={u^2-uv+v^2\over uv}={u^2+v^2\over uv}-1=5.41\] This problem is easiest to solve by guessing a few values and seeing that \(x=2\) works because \[{8^2+125^2\over20^2+50^2}={64+15625\over400+2500}={15689\over2900} ={29\cdot541\over29\cdot100}=5.41\] Proving this is the only answer is more complicated, one possible way is showing the function increases for \(x>0\). By plotting it in a graphing tool, the function is even and the solution can be seen easily. The function can be shown to be even algebraically.
It's not stated, but this is a regular hexagon and the triangle at the top is equilateral. Notice how the height of the triangle is divided in 2 so the area \(x\) is half the area of that triangle. Given side length \(s\), an equilateral triangle has area \(s^2\sqrt{3}/4\). Using the provided \(s\), the area of the shaded area is \[{1\over2}{\left(2\sqrt{2}\sqrt[4]{3}\right)^2\sqrt{3}\over4} ={1\over8}\left(4\cdot2\cdot\sqrt{3}\right)\sqrt{3}=3\]
\(x\) and \(y\) are natural numbers such that \(8x+7y=53\)
The simplest way is to try different \(x\) until \(8x\) exceeds 53. Another way is observing that \(53\equiv5\pmod{8}\). Then \[53-7y\equiv5+y\pmod{8}\] So we need \(y=3,11,\ldots\) for the quantity \(53-7y\) to be divisible by 8. The positive number restriction means only \(y=3\) is suitable which gives us \[8x+7\cdot3=53\Rightarrow 8x=53-21=32\Rightarrow x=4\]
The torque applied is the mass times the distance from the center. For each of these blocks, we use the midpoint. The values to balance form the equation \[7\cdot2.5+5\cdot1.5=1\cdot2.5+x\cdot4.5\Rightarrow17.5+7.5=2.5+4.5x \Rightarrow22.5=4.5x\Rightarrow x=5\]
Find the sum of the roots of \(x^8-7x^7+8\)
Using Vieta's formula, is is \(-a_7/a_8=-(-7)/1=7\).
\(y\) is a positive integer and \(x=1+y+y^2+\ldots+y^7\) is a power of a prime.
By finite geometric sum, \(x={y^8-1\over y-1}\) when \(y>1\). This can be further factored: \[\begin{align}&x={y^8-1\over y-1}={(y^4+1)(y^4-1)\over y-1} ={(y^4+1)(y^2+1)(y^2-1)\over y-1}\\& ={(y^4+1)(y^2+1)(y+1)(y-1)\over y-1}=(y^4+1)(y^2+1)(y+1)\end{align}\] From here, \(y+1\) is a power of a prime. Suppose \(y+1=p^a\) for some prime \(p\) and integer \(a>0\). Then \[y=p^a-1\Rightarrow y^2=p^{2a}-2p^a+1\Rightarrow y^2+1=p^{2a}-2p^a+2\] This quantity must be equal to some other power of the same prime so suppose for some integer \(b>0\) \[p^{2a}-2p^a+2=p^b\Rightarrow 2=p^b-p^{2a}+2p^a\] We see that 2 is a multiple of \(p\) therefore our prime must be \(p=2\). This means \(y+1=2^a\). Then \[y^2=(2^a-1)^2=2^{2a}-2\cdot2^a+1\Rightarrow y^2+1=2^{2a}-2\cdot2^a+2=2^b\] The only time this is true is when \(a=1\). We can see this by factoring it to \[y^2+1=2\cdot(2^{2a-1}-2^a+1)\] which is twice an odd number whenever \(a>1\). Therefore, the only possibility is \(y+1=2\Rightarrow y=1\). In this case, \(x=8=2^3\) which is the solution.
Find the number of factors of 36.
Factor it to \(36=2^2\cdot3^2\). For each prime, there are 3 power choices to form a factor (0,1,2). Therefore the number of factors is \(3\times3=9\).
\[x{42-30+27-35\over45}={40\over45}\Rightarrow 4x=40\Rightarrow x=10\]
We should focus on the first equation to find what \(\theta\) is. This might make us think of the 3-4-5 right triangle which is related. That can be used to find \(\sin(\theta)=3/5,\cos(\theta)=4/5\). These values can be used to find \(x\) in the second equation \[5\sin\theta+5\cos\theta+3\cot\theta=5(3/5)+5(4/5)+3{4/5\over3/5}=3+4+4=11\]
Multiply by \(2/2\) so the denominator becomes \(2^x\) \[2\sum_{x=0}^\infty{2x+1\over2^x}=2\left[2\sum_{x=0}^\infty{x\over2^x} +\sum_{x=0}^\infty2^{-x}\right]\] The simpler sum is \[\sum_{x=0}^\infty2^-x={1\over1-1/2}={1\over1/2}=2\] The other sum can be rewritten as a double summation by changing the order of summation thinking about the following table \[\begin{array}{cccc}1/2&&&\\1/4&1/4&&\\1/8&1/8&1/8&\\\vdots&&&\ddots\\ \end{array}\] Each term corresponds to a row so now if we sum column by column \[\begin{align}&\sum_{x=1}^\infty{x\over2^x}=\sum_{x=1}^\infty \sum_{y=x}^\infty{2^{-y}}=\sum_{x=1}^\infty\left[\sum_{y=0}^\infty2^{-y} -\sum_{y=0}^{x-1}2^{-y}\right]=\sum_{x=1}^\infty\left[ {1\over1-1/2}-{1-2^{-x}\over1-1/2}\right]\\& =\sum_{x=1}^\infty(2-2+2\cdot2^{-x})=2\sum_{x=1}^\infty2^{-x} =2\left[{1\over1-1/2}-2^0\right]=2\end{align}\] So it turns out both of these summations are 2 and if we substitute them into the expression we found earlier, the answer is \(6\).
Find the largest solution of \((25-2y)^{(y^2-25)}=1\).
If \(|25-2y|\neq1\) then we need \(y^2-25=0\Rightarrow y=\pm5\).
If \(|25-2y|=1\) then \(y=12,13\). The only requirement is the exponent is positive to avoid the division by zero problem, which is true for these values. Therefore, the largest solution is \(y=13\).
Two natural numbers differ by 4. The sum of their squares is 106. Find their sum.
Without loss of generality, let \(a,b\) be the 2 numbers with \(a>b\). Then \(a-b=4\) and \(a^2+b^2=106\). By substitution \[(b+4)^2+b^2=106\Rightarrow2b^2+8b+16=106\Rightarrow b^2+4b-45=0 \Rightarrow(b+9)(b-5)=0\] The solutions are \(b=5,-9\) so the only natural number solution is \(b=5\). Then \(a=b+4=9\) so the sum is \(a+b=14\).
The easiest way to solve this is probably by first finding \(\tan(\pi/12)\) which can be done with the angle difference identity \[\tan(\pi/3-\pi/4)={\tan(\pi/3)-\tan(\pi/4)\over1+\tan(\pi/3)\tan(\pi/4)} ={\sqrt{3}-1\over1+\sqrt{3}}\] Then from this we can easily find \[\cot(\pi/12)={1+\sqrt{3}\over\sqrt{3}-1}\] These can be substituted in with some gross algebra \[\begin{align}&{\cot^3(\pi/12)-\tan^3(\pi/12)\over2\sqrt{3}} ={\left({1+\sqrt{3}\over\sqrt{3}-1}\right)^3 -\left({\sqrt{3}-1\over1+\sqrt{3}}\right)^3\over2\sqrt{3}} ={{6\sqrt{3}+10\over6\sqrt{3}-10}-{6\sqrt{3}-10\over6\sqrt{3}+10} \over2\sqrt{3}}\\& ={\left(6\sqrt{3}+10\right)^2-\left(6\sqrt{3}-10\right)^2 \over2\sqrt{3}\left(6\sqrt{3}-10\right)\left(6\sqrt{3}+10\right)} ={36\cdot3+100+120\sqrt{3}-36\cdot3-100+120\sqrt{3} \over2\sqrt{3}\left(36\cdot3-100\right)} ={240\sqrt{3}\over16\sqrt{3}}=15\end{align}\]
This integral does not seem solvable with any standard integration techniques and the result from a computer algebra system is very complicated. We should somehow be able to find \(y=3\pi/16\). Below is code to approximate the solution and guess what fraction it is assuming it is supposed to be a multiple of \(\pi\). This is of course not a proof but a way to get some ideas. As of now, I cannot find a better way to solve this problem.
from fractions import Fraction
from math import pi
f = lambda t : (t**4)/(t**4-t**2+1)**4
dt = 0.001
stop = 5.0
area = 0.0
i = 0
while True:
x1 = i*dt
x2 = (i+1)*dt
h1 = f(x1)
h2 = f(x2)
if x1 >= stop:
break
area += (h1 + h2) * dt / 2
i += 1
print(area)
print(Fraction(area/pi).limit_denominator())
The code sums trapezoids of width \(0.001\) on \([0,5]\), a total of 5000 trapezoids. The resulting area is \(\approx0.5890486\) and if divided by \(\pi\) it is very close to \(3/16\). At \(t=5\), the function is already below \(10^{-9}\) and decreases asymptotically like \(t^{-12}\).
Find the size of the internal that satisfies \(|2x-5|-7\leq10\)
Add 7 to each side \[|2x-5|\leq17\] Use the absolute value equation \[-17\leq2x-5\leq17\] Add 5 \[-12\leq2x\leq22\] Divide 2 \[-6\leq x\leq11\] The interval size is \(|11-(-6)|=17\)
The horizontal side length of the rectangle is \(s=3\sqrt[4]{2}\). To find the height, form a right triangle with the line connecting the centers of the 2 circles. \[(s/2)^2+h^2=(3s/2)^2\Rightarrow h^2=(9/4-1/4)s^2=2s^2\] Therefore, \(h=\sqrt{2}s\) so the area of the rectangle is \[sh=3\sqrt[4]{2}\cdot\sqrt{2}\cdot3\sqrt[4]{2}=3\cdot3\cdot2=18\]
Each edge of a cube is increased by \(6\%\). The volume is thus increased by \(x\%\) (round).
If the old edge is \(1\), then the edge length would be \(1.06\). We can approximate the volume \[(1+0.06)^3=1+3\cdot0.06+3\cdot0.06^2+0.06^3 =1+0.18+3\cdot0.0036+0.06^3\approx1+0.18+0.01=1.19\] The volume has increased by \(19\%\). The exact value is \(19.1016\%\).
We can first simplify the expression for \(x\) a bit \[x=16\sin^2\left({\pi\over10}\right)+8\sin\left({\pi\over10}\right)+16\] By double angle identity \[=8\left(1-\cos\left({\pi\over5}\right)\right) +8\sin\left({\pi\over10}\right)+16 =24+8\left(\sin\left({\pi\over10}\right)-\cos\left({\pi\over5}\right)\right)\] Next we will find \(\cos(\pi/5)\). Let \(\theta=\pi/5\). Start with the identity \[\sin(2\theta)=\sin(\pi-2\theta)\Rightarrow\sin(2\theta)=\sin(3\theta)\] On the left we have \[\sin(\theta+\theta)=2\sin(\theta)\cos(\theta)\] On the right we have \[\sin(2\theta+\theta)=\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) =2\sin(\theta)\cos^2(\theta)+\sin(\theta)(1-2\sin^2(\theta))\] \[=2\sin(\theta)(1-\sin^2(\theta))+\sin(\theta)-2\sin^3(\theta) =3\sin(\theta)-4\sin^3(\theta)\] Next write the equation and divide by \(\sin(\theta)\neq0\) \[3\sin(\theta)-4\sin^3(\theta)=2\sin(\theta)\cos(\theta) \Rightarrow3-4\sin^2(\theta)=2\cos(\theta)=3-4(1-\cos^2(\theta))\] So in terms of \(\cos(\theta)\) we have \[4\cos^2(\theta)-2\cos(\theta)-1=0\] By quadratic formula \[\cos(\theta)={2\pm\sqrt{20}\over8}={1\pm\sqrt{5}\over4}\] We must take the positive solution since \(\cos(\theta)>0\). Therefore, \[\cos\left({\pi\over5}\right)={1+\sqrt{5}\over4}\] We can find the other quantity needed with the half angle identity \[\sin\left({\pi\over10}\right)=\pm\sqrt{1-\cos\left({\pi\over5}\right)\over2} =\pm\sqrt{3-\sqrt{5}\over8}\] We also take the positive solution since \(\sin(\theta/2)>0\) Then \[\begin{align}x&=24+8\left(\sqrt{3-\sqrt{5}\over8}-{1+\sqrt{5}\over4}\right) =24+8\left(\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right) \over8\left(3+\sqrt{5}\right)}-{1+\sqrt{5}\over4}\right)\\& =24+8\left(\frac{1}{\sqrt{6+2\sqrt{5}}}-\frac{1+\sqrt{5}}{4}\right) =24+8\left(\frac{1}{\sqrt{\left(1+\sqrt{5}\right)^2}} -\frac{1+\sqrt{5}}{4}\right)\\&=24+8\left({1\over1+\sqrt{5}} -{1+\sqrt{5}\over4}\right)=24+8\left({1-\sqrt{5}\over-4} +{1+\sqrt{5}\over-4}\right)=24+8\left(2\over-4\right)=20\end{align}\]
Find the count of integers where \((n^2-101)\) is negative.
\[n^2-101<0\Rightarrow n^2<101\Rightarrow-\sqrt{101}<n<\sqrt{101}\] Since \(n\) is an integer, this bound can become \(-10\leq n\leq10\) so there are 21 solutions.
In 2005, Alex is four times as old as Beth. Cal is two times as old as Beth, and Alex is four years older than Cal. How old is Beth in 2025?
In 2005, \(A=4B,C=2B,A=C+4\). Using this information: \[A=4B=2C=2(A-4)=2A-8\Rightarrow A=8\] Therefore, \(B=2,C=4\). We are looking for \(B+20=22\) since it asks for 20 years later in 2025.
The sum of \(x\) consecutive numbers is \(529\). Their average is \(x\).
For the average, we have \[{529\over x}=x\Rightarrow x^2=529\Rightarrow x=\pm23\] Since \(x>0\), we must have \(x=23\). The fact that they are consecutive was not important to solve this. If we were to solve for that, we would find the consecutive numbers are \(12,13,\ldots,34\).
How many digits are in the largest left-truncatable prime?
There are lists of these published on the internet. Out of the 4260 base 10 left-truncatable primes, the largest is 357686312646216567629137 which is 24 digits. They can be computed quickly with some code. Below is an example in Python using the gmpy2 package which has a probable prime test (it turns out the probable test is correct for all the numbers tested in this case). It also finds the maximum right truncatable prime which is a similar process.
import gmpy2
def left(n:str):
if n == '' or gmpy2.is_prime(int(n)): # type:ignore
if n != '':
yield int(n)
for i in '123456789':
yield from left(i+n)
left_list = list(left(''))
def right(n:str):
if n == '' or gmpy2.is_prime(int(n)): # type:ignore
if n != '':
yield int(n)
for i in '123456789':
yield from right(n+i)
right_list = list(right(''))
print(len(left_list),max(left_list),len(str(max(left_list))))
print(len(right_list),max(right_list),len(str(max(right_list))))
Find the sum of the digits of any three-digit number times 999
TODO use paper
Zia used \(x\) five dollar bills to buy books. He could have used 21 fewer 20 dollar bills instead.
This tells us the total cost was \(5x\) dollars. If \(x-21\) 20 dollar bills are used, then the total cost is \(20(x-21)\). This gives us an equation to solve \[5x=20(x-21)\Rightarrow5x=20x-420\Rightarrow15x=420\Rightarrow x=28\]
This is the way to write the number 29 in Chinese.
Kiran can paint half a room in 1 hour, and Jake can paint two-thirds of a room in 2 hours. How long will it take them to point 25 rooms?
This tells us that Kiran paints 0.5 rooms per hour and Jake paints 1/3 of a room every hour. The total rate is \(1/2+1/3=5/6\) rooms per hour. Therefore it will take them \(25/(5/6)=30\) hours to paint 25 rooms.
There are 50 integers to consider. We need to subtract away how many are divisible by the square of a prime. For \(2^2=4\) it is \(104,108,\ldots,148\) so 12. For \(3^2=9\) it is 5. For \(5^2=25\) is it 2. For \(7^2=49\) it is 1. For \(11^2=121\) it is also 1. We can stop since \(13^2>150\). The total is now \[50-12-5-2-1-1=29\] By the inclusion-exclusion principle, we must add back those which were subtracted twice so the numbers divisible by the squares of 2 primes. Only \(2^2\cdot3^2=36\) ends up being counted to add back 2 numbers. Continuing the inclusion-exclusion principle further is not necessary since the smallest would be \(2^2\cdot3^2\cdot5^2=900>150\). Therefore the size of this set is \(29+2=31\). If we were to solve this by testing numbers individually, we would find the following list of 31 numbers: 101, 102, 103, 105, 106, 107, 109, 110, 111, 113, 114, 115, 118, 119, 122, 123, 127, 129, 130, 131, 133, 134, 137, 138, 139, 141, 142, 143, 145, 146, 149.