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## Circular arc | AMC 10A ,2012 | Problem No 18

Try this beautiful Problem on Geometry from Circular arc from (AMC 10 A, 2012).

## Circular arc – AMC-10A, 2012- Problem 18

The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2 \pi}{3},$ where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side $2 .$ What is the area enclosed by the curve?

,

• $2 \pi+6$
• $2 \pi+4 \sqrt{3}$
• $3 \pi+4$
• $2 \pi+3 \sqrt{3}+2$
• $\pi+6 \sqrt{3}$

Geometry

Circle

Hexagon

## Suggested Book | Source | Answer

Pre College Mathematics

AMC-10A, 2012

#### Check the answer here, but try the problem first

$\pi+6 \sqrt{3}$

## Try with Hints

#### First Hint

We have to find out the area enclosed by the curve. but we can not find out eassily. Now we join the centre of 9 congruent circles. it will form a hexagon.

Can you find out the required area of the closed curve? Can you finish the problem?

#### Second Hint

The area of the Hexagon = $(6)(0.5)(2 \sqrt{3})=6 \sqrt{3}$. But we have to find out the area area enclosed by the curve. Then add the areas of the three sectors outside the hexaqon and subtract the areas of the three sectors inside the hexagon but outside the fiqure to get the area enclosed in the curved figure

#### Third Hint

The areas of the three sectors outside the hexaqon=$2 \pi$ .

the areas of the three sectors inside the hexagon but outside the figure $(\pi)$

Therefore the area enclosed by the curve is $\pi+6 \sqrt{3}$

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## Area of rectangle | AMC 10A ,2012 | Problem No 21

Try this beautiful Problem on Geometry from Area of rectangle from (AMC 10 A, 2012).

## Area of rectangle – AMC-10A, 2012- Problem 21

Let points $A=(0,0,0), B=(1,0,0), C=(0,2,0),$ and $D=(0,0,3)$. Points $E, F, G,$ and $H$ are midpoints of line segments $\overline{B D}, \overline{A B}, \overline{A C},$ and $\overline{D C}$ respectively. What is the area of rectangle $E F G H ?$

,

• $\sqrt{2}$
• $\frac{2 \sqrt{5}}{3}$
• $\frac{3 \sqrt{5}}{4}$
• $\sqrt{3}$
• $\frac{2 \sqrt{7}}{3}$

### Key Concepts

Tetrahedron

Area of rectangle

Co -ordinate geometry

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2012, Problem 21

#### Check the answer here, but try the problem first

$\frac{3 \sqrt{5}}{4}$

## Try with Hints

#### First Hint

We have to find out the area of the rectangle $EFGH$. so we have to compute the co-ordinate of the points $E$, $F$, $G$, $H$ . Next we have to find out the length of the sides $EF$, $FG$ , $GH$, $EH$. Next since rectangle area will be $EF$ $\times FG$

Can you solve the problem?

#### Second Hint

Now co-ordinates of the points are $E(0.5,0,1.5), F(0.5,0,0), G(0,1,0), H(0,1,1.5)$. The vector $E F$ is (0,0,-1.5) , while the vector $H G$ is also (0,0,-1.5) , meaning the two sides $E F$ and $G H$ are parallel. Similarly, the vector $F G$ is (-0.5,1,0) , while the vector $E H$ is also (-0.5,1,0) . Again, these are equal in both magnitude and direction, so $F G$ and $E H$ are parallel. Thus, figure $E F G H$ is a parallelogram.The area of the Hexagon = $(6)(0.5)(2 \sqrt{3})=6 \sqrt{3}$. But we have to find out the area area enclosed by the curve. Then add the areas of the three sectors outside the hexaqon and subtract the areas of the three sectors inside the hexagon but outside the fiqure to get the area enclosed in the curved figure

Can you solve the problem?

#### Third Hint

Taking the dot product of vector $E F$ and vector $F G$ gives $0 \cdot-0.5+0 \cdot 1+-1.5 \cdot 0=0,$ which means the two vectors are perpendicular. (Alternately, as above, note that vector $E F$ goes directly down on the z-axis, while vector $F G$ has no z-component and lie completely in the xy plane.) Thus, the figure is a parallelogram with a right angle, which makes it a rectangle.

Using the distance formula we get $E F=\frac{3}{2} \text { and } F G=\frac{\sqrt{5}}{2}$

Therefore area of the rectangle $EFGH$=$EF \times GH$=$\frac{3}{2} \cdot \frac{\sqrt{5}}{2}$=$\frac{3 \sqrt{5}}{4}$

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# Understand the problem

[/et_pb_text][et_pb_text _builder_version=”3.27″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ box_shadow_style=”preset2″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” _i=”1″ _address=”0.0.0.1″]Positive real numbers $a$ and $b$ have the property that$$\sqrt{\log{a}} + \sqrt{\log{b}} + \log \sqrt{a} + \log \sqrt{b} = 100$$and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is $ab$? $\textbf{(A) } 10^{52} \qquad \textbf{(B) } 10^{100} \qquad \textbf{(C) } 10^{144} \qquad \textbf{(D) } 10^{164} \qquad \textbf{(E) } 10^{200}$

# 2019 AMC 12A Problems/Problem 15

[/et_pb_text][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”3.27″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff” _i=”2″ _address=”0.1.0.2″][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″ _i=”0″ _address=”0.1.0.2.0″]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”3.27″ hover_enabled=”0″ _i=”1″ _address=”0.1.0.2.1″]Given both $\sqrt {\log a} , \sqrt {\log b}$ are positive integers .  $\Rightarrow$ both $\log a ,\log b$ are perfect squares . similarly , both $\log {\sqrt a} , \log {\sqrt b}$ are positive integers. $\Rightarrow$ both $a ,b$ are perfect squares .[/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”3.27″ hover_enabled=”0″ _i=”2″ _address=”0.1.0.2.2″]so $$\log a = m^2 , \log b = n^2$$ where $m,n \in {Z^+}$ $\Rightarrow a= 10^{m^2} , b= 10^{n^2}$  and as both $a, b$ are perfect squares  $\\ \Rightarrow 10^{m^2} ,10^{n^2}$ are both perfect squares i.e $10^{m^2} = p^2,10^{n^2} =q^2$ , where $p,q \in {Z^+}$ . $\\ \Rightarrow \frac {m^2}{2} , \frac {n^2}{2}$ are integers $\\ \Rightarrow 2|m^2 , 2|n^2$ $\\ \Rightarrow 2|m, 2|n$ (as $2$ is a prime number ) $\\$ so now $a= 10^{4x^2} , b= 10^{4y^2}$ can be put in the original equation , where  $x,y \in {Z^+}$ .  [/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”3.27″ hover_enabled=”0″ _i=”3″ _address=”0.1.0.2.3″]  Now to get the solution from the derived equation i.e. $2x + 2y + 2x^2 + 2y^2 =100$  multiply both the sides by $2$ and then add $2$ in both sides to arrive at $(2x+1)^2 + (2y+1)^2 = 202$ .[/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”3.27″ hover_enabled=”0″ _i=”4″ _address=”0.1.0.2.4″]Now use trial and error method to express $202$ as a sum of two odd perfect squares . Finally the only way i.e. $9^2 + 11^2 = 202$ .  So without loss of generality it can be written that $(2x+1) = 9 , (2y+1)= 11$ So $a= 10^{64} , b= 10^{100}$  and $ab = 10^{164}$[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version=”3.22.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″ min_height=”12px” custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” _i=”7″ _address=”0.1.0.7″]