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## Area of cube’s cross section |Ratio | AMC 8, 2018 – Problem 24

Try this beautiful problem from Geometry: Ratio of the area of cube’s cross section . You may use sequential hints to solve the problem.

## Area of cube’s cross section – AMC-8, 2018 – Problem 24

In the cube ABCDEFGH with opposite vertices C and E ,J and I are the mid points of segments FB and HD respectively .Let R be the ratio of the area of the cross section EJCI to the area of one of the faces of the cube .what is $R^2$ ?

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

### Key Concepts

Geometry

Area

Pythagorean theorem

Answer:$\frac{3}{2}$

AMC-8(2018) Problem 24

Pre College Mathematics

## Try with Hints

EJCI is a rhombus by symmetry

Can you now finish the problem ……….

Area of rhombus is half product of its diagonals….

can you finish the problem……..

Let Side length of a cube be x.

then by the pythagorean  theorem$EC=X \sqrt {3}$

$JI =X \sqrt {2}$

Now the area of the rhombus is half product of its diagonals

therefore the area of the cross section is $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$

This shows that $R= \frac{\sqrt6}{2}$

i.e$R^2=\frac{3}{2}$

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## Radius of a Semi Circle -AMC 8, 2017 – Problem 22

Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.

## AMC-8(2017) – Geometry (Problem 22)

In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?

• $\frac{7}{6}$
• $\frac{10}{3}$
• $\frac{9}{8}$

### Key Concepts

Geometry

congruency

similarity

Answer:$\frac{10}{3}$

AMC-8(2017)

Pre College Mathematics

## Try with Hints

Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.

Can you now finish the problem ……….

Now the $\triangle ODB$and $\triangle OCB$ are congruent

can you finish the problem……..

Let x be the radius of the semi circle

Now the $\triangle ODB$ and $\triangle OCB$ we have

OD=OC

OB=OB

$\angle ODB$=$\angle OCB$= 90 degree`

so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)

BD=BC=5

And also $\triangle ODA$ and $\triangle BCA$ are similar….

$\frac{8}{12}$=$\frac{x}{5}$

i.e x =$\frac{10}{3}$

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# Understand the problem

[/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″ _i=”1″ _address=”0.0.0.1″]Find all the real Polynomials P(x) such that it satisfies the functional equation: $P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x$.

[/et_pb_text][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”3.29.2″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff” hover_enabled=”0″ _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=”3.29.2″ _i=”1″ _address=”0.1.0.2.1″]Well, it is really good that the information polynomial is given! You should use that. What is the first thing that you check in a Polynomial Identity? Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same $n^2$.  But did you observe something fishy?  [/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”3.29.2″ _i=”2″ _address=”0.1.0.2.2″]Now rewrite the equation as $P(2P(x)) - 2P(P(x)) = P(x)^{2}$. Do the Degree trick now… You see it right? Yes, on the left it is $n^2$ and on the RHS it is $2n$. So, there are two cases now… Figure them out!

[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”3.29.2″ _i=”3″ _address=”0.1.0.2.3″]

Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function. Case 2: $P(2P(x)) - 2P(P(x))$ has coefficient zero till $x^2n$. We will study case 1 now. Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function. $P(2P(x)) - 2P(P(x)) = P(x)^{2}$ = $P(2y) - 2P(y) = y^{2}$ where $y = P(x)$. Now, expand using $P(x) = ax^2 + bx +c$, it gives $2ay^2 -c = y^2$… Now find out all such polynomials satisfying this property. For e.g. $\frac{x^2}{2}$ is a solution. If P(x) is constant, prove that $P(x) = 0 / \frac{-1}{2}$.  [/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”3.29.2″ _i=”4″ _address=”0.1.0.2.4″]Case 2: $P(2y) - 2P(y) = y^{2}$. Assume a general form of P(x) = $latex$and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS.  Now, we have already solved it for quadratic or less degree.  [/et_pb_tab][et_pb_tab title=”Techniques Revisited” _builder_version=”3.29.2″ _i=”5″ _address=”0.1.0.2.5″]

• Always Compare the Degree of Polynomials in identities like this. It provides a lot of information.
• Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.
[/et_pb_tab][et_pb_tab title=”Food for Thought” _builder_version=”3.29.2″ hover_enabled=”0″ _i=”6″ _address=”0.1.0.2.6″]
• Find all polynomials $P(2P(x)) - 8P(P(x)) = P(x)^{2}$.
• Find all polynomials $P(cP(x)) – d P(P(x)) = P(x)^{2}$ depending on the values of c and d.

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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″]

# Similar Problems

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## How do I involve my child in challenging mathematics?

“How do I involve my child in challenging mathematics? He gets good marks in school tests but I think he is smarter than school curriculum.”

“My daughter is in 4th grade. What competitions in mathematics and science can she participate in? How do I help her to perform well in those competitions?”

“I have a 6 years old kid. He hates math. How do I change that?”

We often get queries and requests like these from parents around the world. Literally. In fact, the first one came from Oregon, United States, the second one from Cochin, India, and the last one from Singapore.

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## Initiating a child into the world of Mathematical Science

“How do I involve my son in challenging mathematics? He gets good marks in school tests but I think he is smarter than school curriculum.”

“My daughter is in 4th grade. What competitions in mathematics and science can she participate? How do I help her to perform well in those competitions?”

“I have a 6 years old kid. He hates math. How do I change that?”

We often get queries and requests like these from parents around the world. Literally. In fact the first one came from Oregon, United States, second one from Cochin, India and last one from Singapore.