Tag: AMC

Pre College Mathematics

AMC-10A, 2015 Problem-15

$1$

According to the questation we can write $\frac{x+1}{y+1}=\frac{11 x}{10 y}$

\(\Rightarrow xy +11x-10y=0\)

\(\Rightarrow (x-10)(y-11)=-110\)

Now can you finish the problem?

Here \(x\) and \(y\) must positive, so $x>0$ and $y>0$, so $x-10>-10$ and $y+11>11$

Now we have to find out the factors of \(110\) and find out the possible pairs to fulfill the condition….

Now Can you finish the Problem?

uses the factors of $110$ , we can get the factor pairs: $(-1,110),(-2,55),$ and $(-5,22)$
But we can’t stop here because $x$ and $y$ must be relatively prime.
$(-1,110 )$ gives $x=9$ and $y=99.9$ and 99 are not relatively prime, so this doesn’t work.
$(-2,55 )$ gives $x=8$ and $y=44$. This doesn’t work.
$(-5,22)$ gives $x=5$ and $y=11$. This does work.

Therefore the one solution exist

Pre College Mathematics

AMC-10A, 2015 Problem-24

$31$

Given that $ ABCD$ is a quadrilateral whose perimeter $p$, right angles at $B$ and $C, A B=2$, and $C D=A D$. we have to find out How many different values of $p<2015$ are possible.

Now draw a perpendicular $AE$ on $CD$ . Let us assume that $BC=AE=x$ Then $CE=2$ and $DE=y-2$

Now can you finish the problem?

Now from the \(\triangle ADE\) we can write $x^{2}+(y-2)^{2}=y^{2}$

\(\Rightarrow x^{2}-4 y+4=0\)

\(\Rightarrow x^2=4(y-1)\), Thus, $y$ is one more than a perfect square.

Therefore the perimeter will be $p=2+x+2 y=2 y+2 \sqrt{y-1}+2$

Now according to the problem $p<2015$

So, $p=2+x+2 y=2 y+2 \sqrt{y-1}+2 <2015$

Now Can you finish the Problem?

Now $y=31^{2}+1=962$ and $y=32^{2}+1=1025$

Here $y=31^{2}+1=962$ is valid but $y=32^{2}+1=1025$ is not. On the lower side, $y=1$ does not work (because $x>0$ ), but $y=1^{2}+1$ does work. Hence, there are $31$ valid $y$ (all $y$ such that $y=n^{2}+1$ for $1 \leq n \leq 31$ )

Therefore the correct answer is $31$

Pre College Mathematics

AMC-10A, 2014 Problem-23

$2:3$

we have to find out the  The ratio of the area of the folded paper to that of the original paper.

At first we have to find out the midpoint of the dotted line. Now draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. Now it will form a triangle . we have to find out the area of the triangle ….

Now can you finish the problem?

Let us assume the width of the paper is $1$  and the length is $\sqrt 3$.Now the side length of the triangle be $\frac{2 \sqrt{3}}{3}$, $\sqrt{\left(\frac{\sqrt{3}}{3}\right)^{2}+1}=\frac{2 \sqrt{3}}{3}$, and $\sqrt{\left(\frac{\sqrt{3}}{3}\right)^{2}+1}=\frac{2 \sqrt{3}}{3}$

Now for the area of the of the Paper,

It is an equilateral triangle with height $\frac{\sqrt{3}}{3} \cdot \sqrt{3}=1$ and the area will be $\frac{\frac{2 \sqrt{3}}{3} \cdot 1}{2}=\frac{\sqrt{3}}{3}$

Therefore the area of the paper will be $ 1 . \sqrt 3=\sqrt 3$

Now Can you finish the Problem?

Now the area of the folded paper is $\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2 \sqrt{3}}{3}$

Therefore the ratio of the area of the folded paper to that of the original paper is $2:3$

Pre College Mathematics

AMC-10A, 2017 Problem-22

$\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$

Given that ABC is a equilateral triangle whose \(AB\) & \(AC\) are the tangents of the circle whose centre is \(O\). We have to find out the fraction of the area of $\triangle A B C$ lies outside the circle

we have to find out thr ratio of the areas of Blue colour : Red colour area. Therefore we have to findout the area of the circle and Triangle ABC.

Later we have to find out red area and subtract from the Triangle ABC.

Now can you finish the problem?

Let the radius of the circle be $r$, and let its center be $O$. since $\overline{A B}$ and $\overline{A C}$ are tangent to circle $O$, then $\angle O B A=\angle O C A=90^{\circ}$, so $\angle B O C=120^{\circ} .$ Therefore, since $\overline{O B}$ and $\overline{O C}$ are equal to $r$, then $\overline{B C}=r \sqrt{3}$. The area of the equilateral triangle is $\frac{(r \sqrt{3})^{2} \sqrt{3}}{4}=\frac{3 r^{2} \sqrt{3}}{4},$ and the area of the sector we are subtracting from it is $\frac{1}{3} \pi r^{2}-\frac{1}{2} r \cdot r \cdot \frac{\sqrt{3}}{2}=\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4} .$

Now Can you finish the Problem?

Therefore the area outside the circle is $\frac{3 r^{2} \sqrt{3}}{4}-\left(\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4}\right)=r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}$

Therefore the Required fraction is $\frac{r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}}{\frac{3 r^{2} \sqrt{3}}{4}}$

Pre College Mathematics

AMC-10A, 2009 Problem-12

$13$

We have to find out \(BD\) . Given that \(A B=5, B C=17, C D=5, D A=9,\). Notice that \(BD\) divides quadrilateral $A B C D $ into \(\triangle ABD\) and \(\triangle BCD\). Can you find out using Triangle inequality……?

We know that sum of the two sides of a triangle is greater than the Third side and difference between two sides is lesser than the third side

By the triangle inequality we have \(BD \text { < }DA+AB=9+5=14\) and \(BD \text { > }BC-CD=17-5=12\)

Therefore We got that $12< B D< 14$, and as we know that $B D$ is an integer, we must have $B D$=\(13\)

Pre College Mathematics

AMC-10A, 2009 Problem-23

$6$

Given that Convex quadrilateral $A B C D$ has $A B=9$ and $C D=12$. Diagonals $A C$ and $B D$ intersect at $E, A C=14$, and $\triangle A E D$ and $\triangle B E C$ have equal areas. we have to find out the length of \(AE\).

Now if we can show that \(\triangle AEB\) and \(\triangle DEC\) are similar then we can find out \(AE\)?

Can you find out?

Given that area of \(\triangle AED\) and area of \(\triangle BEC\) are equal. Now area of \(\triangle ABD\) = area of \(\triangle AED\) + \(\triangle ABE\)

Area of \(\triangle ABC\) = area of \(\triangle AEB\) + \(\triangle BEC\)

Therefore area of \(\triangle ABD\)= area of \(\triangle ABC\) [as area of \(\triangle AED\) and area of \(\triangle BEC\) are equal]

Since triangles $A B D$ and $A B C$ share a base, they also have the same height and thus $\overline{A B} || \overline{C D}$ and $\triangle A E B \sim \triangle C E D$ with a ratio of 3: 4

Can you finish the problem?

Therefore $A E=\frac{3}{7} \times A C,$ so $A E=\frac{3}{7} \times 14=6$