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## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|Suppose that the original circle has radius R

Thus, the circumference of this circle is $2 \pi R$.
When this circle is cut into two pieces and each piece is curled to make a cone, the ratio of the lateral surface areas of these cones is $2: 1$.
This means that the ratio of the areas of the two pieces into which the circle is cut is $2: 1$ since it is these pieces that become the lateral surfaces of the cones.
In other words, the sector cut out is $\frac{1}{3}$ of the area of the circle, which means that its central angle is $\frac{1}{3}$ of the total angle around the centre, or $\frac{1}{3} \times 360^{\circ}=120^{\circ}$.
Since the central angles of the two pieces are $240^{\circ}$ and $120^{\circ}$, which are in the ratio $2: 1$, then the circumference is split in the ratio $2: 1$ when the circle is cut.
Since the circumference of the original circle is $2 \pi R$, then the lengths of the pieces of the circumference are $\frac{4}{3} \pi R$ and $\frac{2}{3} \pi R$.
These pieces become the circumferences of the circular bases of the two cones. Since the ratio of the circumference to the radius of a circle is $2 \pi: 1$, then the radii of the bases of the two cones are $\frac{\frac{4}{3} \pi R}{2 \pi}=\frac{2}{3} R$ and $\frac{\frac{2}{3} \pi R}{2 \pi}=\frac{1}{3} R$.
Since the radius of the original circle becomes the slant height of each cone, then the slant height in each cone is $R$.
In a cone, the radius and the height are perpendicular forming a right-angled triangle with the slant height as its hypotenuse.
Therefore, the height of a cone with slant height $R$ and radius $\frac{2}{3} R$ is
$$\sqrt{R^2-\left(\frac{2}{3} R\right)^2}=\sqrt{R^2-\frac{4}{9} R^2}=\sqrt{\frac{5}{9} R^2}=\frac{\sqrt{5}}{3} R$$
Also, the height of a cone with slant height $R$ and radius $\frac{1}{3} R$ is
$$\sqrt{R^2-\left(\frac{1}{3} R\right)^2}=\sqrt{R^2-\frac{1}{9} R^2}=\sqrt{\frac{8}{9} R^2}=\frac{\sqrt{8}}{3} R$$
The volume of a cone with radius $\frac{2}{3} R$ and height $\frac{\sqrt{5}}{3} R$ is $\frac{1}{3} \pi\left(\frac{2}{3} R\right)^2\left(\frac{\sqrt{5}}{3} R\right)$ which equals $\frac{4 \sqrt{5}}{81} \pi R^3$.
The volume of a cone with radius $\frac{1}{3} R$ and height $\frac{\sqrt{8}}{3} R$ is $\frac{1}{3} \pi\left(\frac{1}{3} R\right)^2\left(\frac{\sqrt{8}}{3} R\right)$ which equals $\frac{\sqrt{8}}{81} \pi R^3$. Dividing the first volume by the second, we obtain
$$\frac{\frac{4 \sqrt{5}}{81} \pi R^3}{\frac{\sqrt{8}}{81} \pi R^3}=\frac{4 \sqrt{5}}{\sqrt{8}}=\frac{4 \sqrt{5}}{2 \sqrt{2}}=\frac{4 \sqrt{5} \sqrt{2}}{4}=\sqrt{5} \sqrt{2}=\sqrt{10}$$
Therefore, the ratio of the larger volume to the smaller volume is $\sqrt{10}: 1$.

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2019 Canadian Intermediate Mathematics Contest Solutions

1. (a) In the diagram, $A C$ is the hypotenuse of a right-angled triangle with legs of length 9 and 12.
By the Pythagorean Theorem, $A C^2=9^2+12^2=81+144=225$.
Since $A C>0$, then $A C=\sqrt{225}=15$.
In the diagram, $C B$ is the hypotenuse of a right-angled triangle with legs of length 3 and 4 .
By the Pythagorean Theorem, $C B^2=3^2+4^2=9+16=25$.
Since $B C>0$, then $C B=\sqrt{25}=5$.
(Note that $A C: C B=15: 5=3: 1$.)
(b) Using the given fact, we can compute the ratio by computing the ratio of the differences of $x$-coordinates. We see that $\frac{11-5}{5-1}=\frac{6}{4}=\frac{3}{2}$.
We could also have this with $y$-coordinates: $\frac{2-5}{5-7}=\frac{-3}{-2}=\frac{3}{2}$.
Therefore, the ratio of lengths $G J: J H$ equals $3: 2$.
(c) Solution 1
The difference between the $x$-coordinates of $D(1,6)$ and $E(7,9)$ is $7-1=6$.
This difference is split in the ratio $1: 2$ when it is written as $2+4$.
The difference between the $y$-coordinates of $D(1,6)$ and $E(7,9)$ is $9-6=3$.
This difference is split in the ratio $1: 2$ when it is written as $1+2$.
Since $D$ has coordinates $(1,6)$ and the $x$ – and $y$-coordinates of $E$ are larger than those of $F$, then $F$ should have coordinates $(1+2,6+1)$ or $(3,7)$.
Verifying, using the points $D(1,6), F(3,7), E(7,9)$, we see that
$$\frac{3-1}{7-3}=\frac{2}{4}=\frac{1}{2} \quad \text { and } \quad \frac{7-6}{9-7}=\frac{1}{2}$$
Thus, $F(3,7)$ splits the line segment joining $D(1,6)$ and $E(7,9)$ in the ratio $1: 2$.
Solution 2
Suppose that $F$ has coordinates $(a, b)$.
Since $F(a, b)$ splits $D(1,6)$ and $E(7,9)$ in the ratio $1: 2$, then $\frac{a-1}{7-a}=\frac{1}{2}$ and $\frac{b-6}{9-b}=\frac{1}{2}$.
From $\frac{a-1}{7-a}=\frac{1}{2}$, we obtain $2 a-2=7-a$ and so $3 a=9$ or $a=3$.
From $\frac{b-6}{9-b}=\frac{1}{2}$, we obtain $2 b-12=9-b$ and so $3 b=21$ or $b=7$.
Thus, $F(3,7)$ splits the line segment joining $D(1,6)$ and $E(7,9)$ in the ratio $1: 2$.

# 滑铁卢数学竞赛代考

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|Suppose that the original circle has radius R

$$\sqrt{R^2-\left(\frac{2}{3} R\right)^2}=\sqrt{R^2-\frac{4}{9} R^2}=\sqrt{\frac{5}{9} R^2}=\frac{\sqrt{5}}{3} R$$

$$\sqrt{R^2-\left(\frac{1}{3} R\right)^2}=\sqrt{R^2-\frac{1}{9} R^2}=\sqrt{\frac{8}{9} R^2}=\frac{\sqrt{8}}{3} R$$

$$\frac{\frac{4 \sqrt{5}}{81} \pi R^3}{\frac{\sqrt{8}}{81} \pi R^3}=\frac{4 \sqrt{5}}{\sqrt{8}}=\frac{4 \sqrt{5}}{2 \sqrt{2}}=\frac{4 \sqrt{5} \sqrt{2}}{4}=\sqrt{5} \sqrt{2}=\sqrt{10}$$

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2019 Canadian Intermediate Mathematics Contest Solutions

1. 因此，这个圆的周长是 $2 \pi R$.
2. 当这个圆被切成两片，每片卷曲成一个圆雉体时，这些圆雉体的侧面面积之比为 $2: 1$.
3. 这意味着圆形被切割成的两个部分的面积之比为 $2: 1$ 因为正是这些碎片成为了雉体的侧面。
4. 换句话说，切出的扇区是 $\frac{1}{3}$ 圆的面积，这意味着它的圆心角是 $\frac{1}{3}$ 围绕中心的总角度，或 $\frac{1}{3} \times 360^{\circ}=120^{\circ}$.
5. 由于两片的中心角为 $240^{\circ}$ 和 $120^{\circ}$ ，它们在比率中 $2: 1$ ，然后圆周按比例分割 $2: 1$ 当圆圈被切嗐时。
6. 由于原圆的周长为 $2 \pi R$, 那么圆周段的长度是 $\frac{4}{3} \pi R$ 和 $\frac{2}{3} \pi R$.
7. 由于原圆的半径变成了每个圆雉的斜高，那么每个圆雉的斜高为 $R$.
8. 在圆雉中，半径和高度垂直，形成一个以斜高为斜边的直角三角形。
9. 因此，具有斜高的圆雉的高度 $R$ 和半径 $\frac{2}{3} R$ 是
10. $$11. \sqrt{R^2-\left(\frac{2}{3} R\right)^2}=\sqrt{R^2-\frac{4}{9} R^2}=\sqrt{\frac{5}{9} R^2}=\frac{\sqrt{5}}{3} R 12.$$
13. 此外，具有倾斜高度的圆雉的高度 $R$ 和半径 $\frac{1}{3} R$ 是
14. $$15. \sqrt{R^2-\left(\frac{1}{3} R\right)^2}=\sqrt{R^2-\frac{1}{9} R^2}=\sqrt{\frac{8}{9} R^2}=\frac{\sqrt{8}}{3} R 16.$$
17. 有半径的圆锥的体积 $\frac{2}{3} R$ 和身高 $\frac{\sqrt{5}}{3} R$ 是 $\frac{1}{3} \pi\left(\frac{2}{3} R\right)^2\left(\frac{\sqrt{5}}{3} R\right)$ 这等于 $\frac{4 \sqrt{5}}{81} \pi R^3$.
18. 有半径的圆锥的体积 $\frac{1}{3} R$ 和身高 $\frac{\sqrt{8}}{3} R$ 是 $\frac{1}{3} \pi\left(\frac{1}{3} R\right)^2\left(\frac{\sqrt{8}}{3} R\right)$ 这等于 $\frac{\sqrt{8}}{81} \pi R^3$. 将第一卷除以第二卷，我们得到
19. $$20. \frac{\frac{4 \sqrt{5}}{81} \pi R^3}{\frac{\sqrt{8}}{81} \pi R^3}=\frac{4 \sqrt{5}}{\sqrt{8}}=\frac{4 \sqrt{5}}{2 \sqrt{2}}=\frac{4 \sqrt{5} \sqrt{2}}{4}=\sqrt{5} \sqrt{2}=\sqrt{10} 21.$$
22. 因此，较大体积与较小体积之比为 $\sqrt{10}: 1$.

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