A trigonometric polynomial ( INMO 2020 Problem 2)

The problem

Suppose $P(x)$ is a polynomial with real coefficients satisfying the condition
$$
P(\cos \theta+\sin \theta)=P(\cos \theta-\sin \theta),
$$
for every real $\theta$. Prove that $P(x)$ can be expressed in the form
$$
P(x)=a_0+a_1\left(1-x^2\right)^2+a_2\left(1-x^2\right)^4+\cdots+a_n\left(1-x^2\right)^{2 n},
$$
for some real numbers $a_0, a_1, a_2, \ldots, a_n$ and nonnegative integer (n).

Hint 1

Using a very standard trigometric identity, we can easily convert the following ,
$$
\begin{aligned}
P(\cos \theta+\sin \theta) & =P(\cos \theta-\sin \theta
\Longrightarrow P\left(\sqrt{2} \sin \left(\frac{\pi}{4}+\theta\right)\right) & =P\left(\sqrt{2} \cos \left(\frac{\pi}{4}+\theta\right)\right) \
\Longrightarrow P(\sqrt{2} \sin x) & =P(\sqrt{2} \cos x)
\end{aligned}
$$
⟹ $P(\sqrt{2} \sin x)=P(\sqrt{2} \cos x) \quad$

Assuming,

$\left(\frac{\pi}{4}+\theta\right)=x$ for all reals $x$. So,

$P(-\sqrt{2} \sin (x))=P(\sqrt{2} \sin (-x))=P(\sqrt{2} \cos (-x))=P(\sqrt{2} \cos (x))=P(\sqrt{2} \sin (x))$ for all $x \in \mathbb{R}$. Since $P(x)=P(-x)$ holds for infinitely many $x$, it must hold for all $x$ (since $P(x)$ is a polynomial). so we get that, $P(x)$ is a even polynomial.

Hint 2

$P(\sqrt{2} \cos (x))=P(\sqrt{2} \sin (x))$ implies that
$$
P(t)=P\left(\sqrt{2} \sin \left(\cos ^{-1}(t / \sqrt{2})\right) \text { putting }, x=\cos ^{-1}(t / \sqrt{2})\right.
$$
for infinitely many $t \in[-\sqrt{2}, \sqrt{2}]$.
$$
\sqrt{2} \sin \left(\cos ^{-1}(t / \sqrt{2})\right)=\sqrt{2-t^2} \text { so we get, } P(x)=P\left(\sqrt{2-t^2}\right)
$$

Again as it is a polynomial function we can extend it all $\mathbb{R}$. And we get, $P(x)=P\left(\sqrt{2-x^2}\right)$ for all reals (x)

Hint 3

Since $P(x)$ is even, we can choose an even polynomial $Q(x)$ such that, $Q(x)=P(\sqrt{x+1}) \cdot P(\sqrt{1+x}$=$Q(x)=a_0+a_1 x^2+a_2 x^4+\cdots+a_n x^{2 n}$ now take, $\sqrt{1+x}=y$ and you get the polynomial of required form.

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Kites in Geometry | INMO 2020 Problem 1

Understand the problem

Let $ \Gamma_1 $ and $ \Gamma_2 $ be two circles with unequal radii, with centers $ O_1 $ and $ O_2 $ respectively, in the plane intersecting in two distinct points A and B. Assume that the center of each of the circles $ \Gamma_1 $ and $ \Gamma_2 $ are outside each other. The tangent to $ \Gamma_ 1 $ at B intersects $ \Gamma_2 $ again at C, different from B; the tangent to $ \Gamma_2 $ at B intersects $ \Gamma_1 $ again in D different from B. The bisectors of $ \angle DAB $ and $ \angle CAB $ meet $ \Gamma_1 $ and $ \Gamma_2 $ again in X and Y, respectively. different from A. Let P and Q be the circumcenters of the triangles ACD and XAY, respectively. Prove that PQ is perpendicular bisector of the line segment $ O_1 O_2 $.

Tutorial Problems... try these before watching the video.

1. Suppose $ P O_1 Q O_2 $ be a kite (that is $ PO_1 = PO_2 $ and $ QO_1 1 = QO_2 $. Show that PQ is perpendicular bisector of the other diagonal $ O_1 O_2 $.$.

2. Show that for any two circles intersecting each other at two distinct points, the common chord is bisected perpendicularly by the line joining the center.

You may send solutions to support@cheenta.com. Though we usually look into internal students work, we will try to give you some feedback.

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INMO 2020 Problems, Solutions and Hints

Problems

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6

This is a work in progress. More discussions will be uploaded soon.

INMO 2020 Problem 4

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Let $latex n\ge 3$ be an integer and $latex a_1,a_2,\cdots a_n$ be real numbers satisfying $latex 1<a_2\le a_2\le a_3\cdots \le a_n$. If $latex \Sigma_ia_i=2n$ then prove that $latex 2+a_1+a_1a_2+a_1a_2a_3+\cdots +a_1a_2\cdots a_{n-1}\le a_1a_2\cdots a_n$.

[/et_pb_text][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" admin_label="Blog" _builder_version="3.22" custom_margin="|||" custom_padding="0px|0px|21px|0px|false|false"][et_pb_row _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat" custom_padding="0|0px|24px|0px|false|false" use_custom_width="on" custom_width_px="960px"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_tabs _builder_version="3.12.2"][et_pb_tab title="Hint 1" _builder_version="3.12.2"]

The conditions hint at inequalities involving an order, such as the rearrangement and Chebychev inequalities. Also note that $latex a_i=2$ for all $latex i$ is an equality case, hence we should try to use inequalities in such a way that matches the equality case.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.12.2"]

The RHS can be rewritten as $latex a_1a_2\cdots a_n= a_1a_2\cdots a_n-a_1a_2\cdots a_{n-1}+a_1a_2\cdots a_{n-1}-a_1a_2\cdots a_{n-2}+a_1a_2\cdots a_{n-2}\cdots -a_1+a_1=a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-2}(a_{n-1}-1)+\cdots+ a_1(a_2-1)+a_1$. That is, $latex a_1a_2\cdots a_n-1= a_1a_2\cdots a_n-a_1a_2\cdots a_{n-1}+a_1a_2\cdots a_{n-1}-a_1a_2\cdots a_{n-2}+a_1a_2\cdots a_{n-2}\cdots -a_1+a_1=a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-2}(a_{n-1}-1)+\cdots+ a_1(a_2-1)+(a_1-1)$.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.12.2"]

Now Chebychev inequality gives $latex \frac{a_1a_2\cdots a_n-1}{n}= \frac{a_1a_2\cdot a_{n-1}(a_n-1)+a_1a_2\cdot a_{n-2}(a_{n-1}-1)+\cdots+ a_1(a_2-1)+(a_1-1)}{n}\ge \frac{1+a_1+a_1a_2+\cdots +a_1a_2\cdots a_{n-1}}{n}\cdot\frac{(a_1-1+a_2-1+\cdots +a_n-1)}{n}=\frac{1+a_1+a_1a_2+\cdots +a_1a_2\cdots a_{n-1}}{n}$. Cancelling the denominators, we get the desired result.

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