Math Ga (a) 2019 MEXT Monbukagakusho

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Answered by https://math-gpt.org/

 

(6) Sequence

The sequence is

$$1,\ 4,\ 10,\ 19,\ 31,\ldots$$

Find the differences:

$$\begin{array}{c|ccccc} \text{Term} & 1 & 4 & 10 & 19 & 31\\ \hline \text{First difference} & & 3 & 6 & 9 & 12\\ \text{Second difference} & & & 3 & 3 & 3 \end{array}$$

Since the second difference is constant, the sequence is quadratic.

The first difference before the first term is $0$, so the $n$-th term can be found by adding the differences:

$$a_n=1+3(1+2+\cdots +(n-1)).$$

Using

$$1+2+\cdots +(n-1)=\frac{n(n-1)}{2},$$

we get

$$a_n=1+\frac{3n(n-1)}{2}.$$

Therefore,

$$\boxed{a_n=\frac{3n^2-3n+2}{2}}.$$

For the sum of the first $n$ terms,

$$S_n=\sum_{k=1}^{n}\left(1+\frac{3k(k-1)}{2}\right).$$

Thus,

$$S_n=n+\frac{3}{2}\sum_{k=1}^{n}k(k-1).$$

Since

$$\sum_{k=1}^{n}k(k-1)=\frac{n(n+1)(n-1)}{3},$$

we have

$$S_n=n+\frac{n(n+1)(n-1)}{2}.$$

Simplifying,

$$\boxed{S_n=\frac{n^3+n}{2}}.$$

So the answers for (6) are

$$\boxed{[1\text{-}11]=\frac{3n^2-3n+2}{2}}$$

and

$$\boxed{[1\text{-}12]=\frac{n^3+n}{2}}.$$

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(7) Minimization

We are given

$$\frac{4a+b}{2a}+\frac{4a-3b}{b},$$

where $a,b>0$.

Let

$$x=\frac{b}{a}.$$

Then

$$\frac{4a+b}{2a} =2+\frac{x}{2},$$

and

$$\frac{4a-3b}{b} =\frac{4}{x}-3.$$

Therefore, the expression becomes

$$2+\frac{x}{2}+\frac{4}{x}-3 =\frac{x}{2}+\frac{4}{x}-1.$$

By AM-GM,

$$\frac{x}{2}+\frac{4}{x} \geq 2\sqrt{\frac{x}{2}\cdot\frac{4}{x}} =2\sqrt{2}.$$

Equality occurs when

$$\frac{x}{2}=\frac{4}{x}.$$

Hence,

$$x^2=8,$$

and because $x>0$,

$$x=2\sqrt{2}.$$

Since $x=\frac ba$,

$$\boxed{b=2\sqrt{2}\,a}.$$

The minimum value is

$$2\sqrt{2}-1.$$

Therefore,

$$\boxed{[1\text{-}13]=2\sqrt{2}}$$

and

$$\boxed{[1\text{-}14]=2\sqrt{2}-1}.$$

Solve the above problem.

(9) Divisibility of a base-$6$ number

We have

$$\sum_{k=0}^{n}x_k6^k.$$

For $k\geq 1$,

$$6^k=1+5(1+6+6^2+\cdots+6^{k-1}) =1+5\sum_{l=0}^{k-1}6^l.$$

Therefore,

$$\begin{aligned} \sum_{k=0}^{n}x_k6^k &=x_0+\sum_{k=1}^{n}x_k6^k\\ &=x_0+\sum_{k=1}^{n}x_k \left(1+5\sum_{l=0}^{k-1}6^l\right)\\ &=\boxed{x_0} +\boxed{5}\left(\sum_{k=1}^{n}x_k\sum_{l=0}^{k-1}6^l\right). \end{aligned}$$

Thus,

$$[1\text{-}21]=\boxed{x_0}, \qquad [1\text{-}22]=\boxed{5}.$$

The number is divisible by $5$ exactly when its digits add to a multiple of $5$, because

$$6^k\equiv 1\pmod 5.$$

Hence,

$$\sum_{k=0}^{n}x_k6^k \equiv \sum_{k=0}^{n}x_k\pmod 5.$$

Therefore,

$$[1\text{-}23]=\boxed{5}.$$

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(10) Solving the linear Diophantine equation

We need integer solutions to

$$253x+256y=1.$$

Since $256=253+3$,

$$253x+256y=253(x+y)+3y.$$

Thus,

$$253(x+y)+3y=1.$$

Taking the equation modulo $253$,

$$3y\equiv 1\pmod{253}.$$

Because

$$3(169)=507=2(253)+1,$$

we have

$$y\equiv 169\pmod{253}.$$

So write

$$y=169+253t,$$

where $t$ is an integer. Substituting into the original equation gives

$$x=-171-256t.$$

To minimize $\lvert x\rvert$, choose $t=-1$. Then

$$x=-171+256=85,$$

and

$$y=169-253=-84.$$

Therefore,

$$\boxed{[1\text{-}24]=85}, \qquad \boxed{[1\text{-}25]=-84}.$$

Check:

$$253(85)+256(-84)=21,505-21,504=1.$$

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(11) Translating the parabola

The original function is

$$f(x)=2x^2+3x+1.$$

Translating the graph 2 units to the right replaces $x$ with $x-2$:

$$f(x-2)=2(x-2)^2+3(x-2)+1.$$

Translating it 3 units downward subtracts $3$:

$$y=2(x-2)^2+3(x-2)+1-3.$$

Expand:

$$\begin{aligned} y &=2(x^2-4x+4)+3x-6+1-3\\ &=2x^2-8x+8+3x-8\\ &=2x^2-5x. \end{aligned}$$

Thus,

$$y=2x^2-5x+0,$$

so

$$\boxed{[1\text{-}26]=2}, \qquad \boxed{[1\text{-}27]=-5}, \qquad \boxed{[1\text{-}28]=0}.$$

Solve the above problem.

We are given:

• $D$ lies on $AB$,
• $CD\perp AB$,
• $\angle BAC=\dfrac{\pi}{12}$,
• $AB=2\sqrt{2}$,
• $AD=\sqrt{6}$.

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(1) Find the missing numerator in $\cos\dfrac{\pi}{12}$

Since

$$\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4},$$

use the cosine subtraction formula:

$$\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right) =\cos\frac{\pi}{3}\cos\frac{\pi}{4} +\sin\frac{\pi}{3}\sin\frac{\pi}{4}.$$

Substitute the known values:

$$\cos\frac{\pi}{12} =\frac12\cdot\frac{\sqrt2}{2} +\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}.$$

Thus,

$$\cos\frac{\pi}{12} =\frac{\sqrt2+\sqrt6}{4}.$$

Therefore,

$$\boxed{[2\text{-}1]=\sqrt6}.$$

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(2) Find $AC$

Because $CD\perp AB$, triangle $ACD$ is a right triangle.

At vertex $A$,

$$\cos\frac{\pi}{12} =\frac{AD}{AC}.$$

Therefore,

$$AC=\frac{AD}{\cos(\pi/12)} =\frac{\sqrt6}{(\sqrt6+\sqrt2)/4}.$$

Simplify:

$$AC=\frac{4\sqrt6}{\sqrt6+\sqrt2}.$$

Rationalizing the denominator,

$$AC =\frac{4\sqrt6(\sqrt6-\sqrt2)}{6-2} =\sqrt6(\sqrt6-\sqrt2).$$

Hence,

$$AC=6-2\sqrt3.$$

So,

$$\boxed{[2\text{-}2]=6}.$$

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(3) Find $(BC)^2$

Use the Law of Cosines in triangle $ABC$:

$$BC^2=AB^2+AC^2-2(AB)(AC)\cos\frac{\pi}{12}.$$

Substitute the known values:

$$BC^2=(2\sqrt2)^2+(6-2\sqrt3)^2 -2(2\sqrt2)(6-2\sqrt3)\left(\frac{\sqrt6+\sqrt2}{4}\right).$$

Now simplify each part:

$$(2\sqrt2)^2=8,$$

and

$$(6-2\sqrt3)^2 =36-24\sqrt3+12 =48-24\sqrt3.$$

The product term simplifies to

$$2(2\sqrt2)(6-2\sqrt3)\left(\frac{\sqrt6+\sqrt2}{4}\right) =32\sqrt3.$$

Therefore,

$$BC^2=8+(48-24\sqrt3)-32\sqrt3,$$

so

$$BC^2=56-32\sqrt3.$$

Thus,

$$\boxed{[2\text{-}3]=56}.$$

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(4) Find $BC$

We need to simplify

$$BC=\sqrt{56-32\sqrt3}.$$

Suppose

$$BC=a-2\sqrt6.$$

Then

$$(a-2\sqrt6)^2=a^2+24-4a\sqrt6.$$

We want this to equal

$$56-32\sqrt3.$$

Since $a^2+24=56$,

$$a^2=32,$$

so

$$a=4\sqrt2.$$

Also,

$$4a\sqrt6=4(4\sqrt2)\sqrt6 =16\sqrt{12} =32\sqrt3,$$

which matches the radical term. Therefore,

$$BC=4\sqrt2-2\sqrt6.$$

Hence,

$$\boxed{[2\text{-}4]=4\sqrt2}.$$

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Final answers

$$\boxed{[2\text{-}1]=\sqrt6}$$

$$\boxed{[2\text{-}2]=6}$$

$$\boxed{[2\text{-}3]=56}$$

$$\boxed{[2\text{-}4]=4\sqrt2}$$