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AMC 10 Geometry Math Olympiad USA Math Olympiad

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?

Circular Arc

,

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

Key Concepts


Geometry

Circle

Hexagon

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2012

Check the answer here, but try the problem first

$\pi+6 \sqrt{3}$

Try with Hints


First Hint

Circular arc

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.

Circular arc with hexagon

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

Second Hint

Circular arc with hexagon

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

 hexagon

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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AMC 10 Geometry Math Olympiad USA Math Olympiad

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


Suggested Reading

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

Area of rectangle

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

Area of rectangle

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

Area of rectangle problem

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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USA Math Olympiad

AMC 2019 12A Problem 15 Diophantine Equation

[et_pb_section fb_built=”1″ _builder_version=”3.22.4″ fb_built=”1″ _i=”0″ _address=”0″][et_pb_row _builder_version=”3.25″ _i=”0″ _address=”0.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||” _i=”0″ _address=”0.0.0″][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″ custom_padding=”20px|20px|20px|20px” _i=”0″ _address=”0.0.0.0″]

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 $ab$? $\textbf{(A) }   10^{52}   \qquad        \textbf{(B) }   10^{100}   \qquad    \textbf{(C) }   10^{144}   \qquad   \textbf{(D) }  10^{164} \qquad  \textbf{(E) }   10^{200}$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”3.27″ _i=”1″ _address=”0.1″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||” _i=”0″ _address=”0.1.0″][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”3.27″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px” _i=”0″ _address=”0.1.0.0″][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”3.27″ hover_enabled=”0″ _i=”0″ _address=”0.1.0.0.0″]

2019 AMC 12A Problems/Problem 15

[/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”3.27″ hover_enabled=”0″ _i=”1″ _address=”0.1.0.0.1″ open=”off”]logarithm, diophantine equation[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”3.27″ hover_enabled=”0″ _i=”2″ _address=”0.1.0.0.2″ open=”off”]Medium[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.27″ hover_enabled=”0″ _i=”3″ _address=”0.1.0.0.3″ open=”off”]Mathematical Circles (Russian Experience)[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version=”3.27″ 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″ custom_margin=”48px||48px” custom_padding=”20px|20px|20px|20px” _i=”1″ _address=”0.1.0.1″]

Start with hints

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[/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″]

Connected Program at Cheenta

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url=”https://www.cheenta.com/matholympiad/” url_new_window=”on” button_text=”Learn More” button_alignment=”center” _builder_version=”3.23.3″ custom_button=”on” button_bg_color=”#0c71c3″ button_border_color=”#0c71c3″ button_border_radius=”0px” button_font=”Raleway||||||||” button_icon=”%%3%%” button_text_shadow_style=”preset1″ box_shadow_style=”preset1″ box_shadow_color=”#0c71c3″ background_layout=”dark” _i=”9″ _address=”0.1.0.9″][/et_pb_button][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″ custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” _i=”10″ _address=”0.1.0.10″]

Similar Problems

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USA Math Olympiad

AMC 10 Paper Folding Geometry

We fold a paper using GeoGebra and explore a problem from American Mathematical Contest (AMC 10)

Problem: A rectangular piece of paper whose length is ( \sqrt 3 ) times the width has area A. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area B. What is the ratio B: A?

You may also try another paper folding scenario. Here we make the crease a variable!