Category: Math Olympiad

Pre College Mathematics

AMC-10A, 2018 Problem-23

\(\frac{145}{147}\)

Given that ABC is a right-angle Triangle field . Here The corner at \(B\) is shaded region which is unplanted. now we have to find out fraction of the field is planted?

Now if we join the triangle with the dotted lines then it will be divided into three triangles as shown below…

Therefore there are three triangles . Now if we can find out the area of three triangles and area of the smaller square then it will be eassy to say….

Now can you finish the problem?

Let \(x\) be the side length of the sqare then area will be\(x^2\)

Now area of two thin triangle will be $\frac{x(3-x)}{2}$ and $\frac{x(4-x)}{2}$

area of the other triangle will be \(\frac{1}{2}\times 5 \times 2=5\)

area of the \(\triangle ABC =\frac{1}{2}\times 3 \times 4=6\)

Now Can you finish the Problem?

Therefore we can say that $x^{2}+\frac{x(3-x)}{2}+\frac{x(4-x)}{2}+5=6$

\(\Rightarrow x=\frac{2}{7}\)

Therefore area of the small square will be \(\frac{4}{49}\)

Thererfore our required fraction =Area of the \(\triangle ABC\)-area of the smaller square=\(6- \frac{4}{49}\)=\(\frac{145}{147}\)

Pre College Mathematics

AMC-10B, 2016 Problem-21

$\pi+2$

The given equation is $x^{2}+y^{2}=|x|+|y|$. Expanding this equation we get four equation as mod exist here…

$x^2+y^2-x-y=0$…………………..(1)

$x^2+y^2+x+y=0$………………..(2)

$x^2+y^2-x+y=0$…………………(3)

$x^2+y^2+x-y=0$…………………(4)

using this four equation can you draw the figure ?

Now can you finish the problem?

now four equations can be written as $x^{2}-x+y^{2}-y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}+y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}-x+y^{2}+y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}-y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$ which represents four circles and they overlapping…..

The center of the four circles are $\left(\frac{1}{2}, \frac{1}{2}\right)$, $\left(\frac{-1}{2}, \frac{-1}{2}\right)$,$\left(\frac{1}{2}, \frac{-1}{2}\right)$,$\left(\frac{-1}{2}, \frac{1}{2}\right)$Now we have to find out the region union of the four circles.

Now can you finish the problem?

There are several ways to find the area, but note that if you connect (0,1),(1,0),(-1,0),(0,-1) to its other three respective points in the other three quadrants, you get a square of area 2 , along with four half-circles of diameter $\sqrt{2}$, for a total area of $2+2 \cdot\left(\frac{\sqrt{2}}{2}\right)^{2} \pi=\pi+2$

Pre College Mathematics

AMC-10A, 2016 Problem-19

$20$

Given that , rectangle $A B C D, A B=6$ and $B C=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that Segments $\overline{A E}$ and $\overline{A F}$ intersect $\overline{B D}$ at $P$ and $Q,$ respectively. The ratio $B P: P Q: Q D$ can be written as $r: s: t$. we have to find out $r+s+t ?$, where greatest common factor of \(r,s,t\) is \(1\)

Now $\triangle A P D \sim \triangle E P B$. From this relation we can find out a relation between \(DP\) and \(PB\)

Now can you finish the problem?

Now $\triangle A P D \sim \triangle E P B$\(\Rightarrow\) $\frac{D P}{P B}=\frac{A D}{B E}=3$ Therefore $P B=\frac{B D}{4}$.

SimIarly from the \(\triangle AQD \sim \triangle BQF\) \(\Rightarrow \)$\frac{D Q}{Q B}=\frac{3}{2}$

Therefore we can say that $D Q=\frac{3 \cdot B D}{5}$

Now can you finish the problem?

Therefore $r: s: t=\frac{1}{4}: \frac{2}{5}-\frac{1}{4}: \frac{3}{5}=5: 3: 12,$ so $r+s+t$=\(20\)

Pre College Mathematics

AMC-10A, 2010 Problem-20

$4 \sqrt{2}+4 \sqrt{3}$

Suppose the fly starts from the point \(A\).we have to find out  the maximum possible length. The maximum possible length will be from one corner to another corner such as …..

$A \rightarrow G \rightarrow B \rightarrow H \rightarrow C \rightarrow E \rightarrow D \rightarrow F \rightarrow A$

Now can you find out this maximum path?

Now can you finish the problem?

Given that the side length of the cube is $1$. Therefore the diagonal \(AC\)=\(\sqrt 2\) and the diagonal \(AG=\sqrt 3\). Now we have to find out the path \(AG+GB+BH+HC+CE+ED+DF+FA\).Can you find it ?

Now can you finish the problem?

\(AG+GB+BH+HC+CE+ED+DF+FA\)=\(\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2+\sqrt 3+\sqrt 2\)=(\(4\sqrt3+4\sqrt2\))

Pre College Mathematics

AMC-10A, 2019 Problem-`13

$110^{\circ}$

According to the questation we draw the diagram. we have to find out \(\angle BFC\)

Now \(\angle BEC\) = \(\angle BDC\) =\(90^{\circ}\) (as they are inscribed in a semicircle)

$\angle A C B=40^{\circ} .$ Therefore we can say that \(\angle ABC=70^{\circ}\) (as $\triangle A B C$ be an isosceles triangle with $B C=A C$)

Can you find out the value of $\angle B F C ?$

Now can you finish the problem?

As \(\angle ABC=70^{\circ}\) and \(\angle BEC=90^{\circ}\) Therefore $\angle E C B=20^{\circ}$( as sum of the angles of a triangle is\( 180^{\circ}\)

Similarly $\angle D B C=50^{\circ}$

Now Can you finish the Problem?

Now $\angle B D C+\angle D C B+\angle D B C=180^{\circ} \Longrightarrow 90^{\circ}+40^{\circ}+\angle D B C=180^{\circ} \Longrightarrow \angle D B C$=$50^{\circ}$

$\angle B E C+\angle E B C+\angle E C B=180^{\circ} \Longrightarrow 90^{\circ}+70^{\circ}+\angle E C B=180^{\circ} \Rightarrow \angle E C B$=$20^{\circ}$

we take triangle $B F C$, and find $\angle B F C=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$