Found problems: 3349
2006 China Western Mathematical Olympiad, 3
In $\triangle PBC$, $\angle PBC=60^{o}$, the tangent at point $P$ to the circumcircle$g$ of $\triangle PBC$ intersects with line $CB$ at $A$. Points $D$ and $E$ lie on the line segment $PA$ and $g$ respectively, satisfying $\angle DBE=90^{o}$, $PD=PE$. $BE$ and $PC$ meet at $F$. It is known that lines $AF,BP,CD$ are concurrent.
a) Prove that $BF$ bisect $\angle PBC$
b) Find $\tan \angle PCB$
2013 China Team Selection Test, 2
Let $P$ be a given point inside the triangle $ABC$. Suppose $L,M,N$ are the midpoints of $BC, CA, AB$ respectively and \[PL: PM: PN= BC: CA: AB.\] The extensions of $AP, BP, CP$ meet the circumcircle of $ABC$ at $D,E,F$ respectively. Prove that the circumcentres of $APF, APE, BPF, BPD, CPD, CPE$ are concyclic.
1982 AMC 12/AHSME, 23
The lengths of the sides of a triangle are consescutive integers, and the largest angle is twice the smallest angle. The cosine of the smallest angle is
$\textbf {(A) } \frac 34 \qquad \textbf {(B) } \frac{7}{10} \qquad \textbf {(C) } \frac 23 \qquad \textbf {(D) } \frac{9}{14} \qquad \textbf {(E) } \text{none of these}$
2008 China Team Selection Test, 1
Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \equal{} FG$.
2007 Mediterranean Mathematics Olympiad, 3
In the triangle $ABC$, the angle $\alpha = \angle BAC$ and the side $a = BC$ are given. Assume that $a = \sqrt{rR}$, where $r$ is the inradius and $R$ the circumradius. Compute all possible lengths of sides $AB$ and $AC.$
2004 USAMO, 1
Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
\[
\frac{1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.
\]
When does equality hold?
2008 Singapore Team Selection Test, 1
Let $(O)$ be a circle, and let $ABP$ be a line segment such that $A,B$ lie on $(O)$ and $P$ is a point outside $(O)$. Let $C$ be a point on $(O)$ such that $PC$ is tangent to $(O)$ and let $D$ be the point on $(O)$ such that $CD$ is a diameter of $(O)$ and intersects $AB$ inside $(O)$. Suppose that the lines $DB$ and $OP$ intersect at $E$. Prove that $AC$ is perpendicular to $CE$.
1981 AMC 12/AHSME, 19
In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ and $\theta$ is the measure of $\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals
[asy]
size(230);
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, A=14*dir(36), C=intersectionpoint(B--(9001,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b);
draw(N--B--A--N--M--C--A^^B--M);
markscalefactor=0.1;
draw(rightanglemark(B,N,A));
pair point=N;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$M$", M, S);
label("$N$", N, dir(30));
label("$19$", (A+C)/2, dir(A--C)*dir(90));
label("$14$", (A+B)/2, dir(A--B)*dir(270));
[/asy]
$\displaystyle \text{(A)} \ 2 \qquad \text{(B)} \ \frac{5}{2} \qquad \text{(C)} \ \frac{5}{2} - \sin \theta \qquad \text{(D)} \ \frac{5}{2} - \frac{1}{2} \sin \theta \qquad \text{(E)} \ \frac{5}{2} - \frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$
2005 Baltic Way, 12
Let $ABCD$ be a convex quadrilateral such that $BC=AD$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. The lines $AD$ and $BC$ meet the line $MN$ at $P$ and $Q$, respectively. Prove that $CQ=DP$.
2008 Mathcenter Contest, 3
Let $ABC$ be a triangle whose side lengths are opposite the angle $A,B,C$ are $a,b,c$ respectively. Prove that $$\frac{ab\sin{2C}+bc\sin{ 2A}+ca\sin{2B}}{ab+bc+ca}\leq\frac{\sqrt{3}}{2}$$.
[i](nooonuii)[/i]
1995 IMO Shortlist, 3
The incircle of triangle $ \triangle ABC$ touches the sides $ BC$, $ CA$, $ AB$ at $ D, E, F$ respectively. $ X$ is a point inside triangle of $ \triangle ABC$ such that the incircle of triangle $ \triangle XBC$ touches $ BC$ at $ D$, and touches $ CX$ and $ XB$ at $ Y$ and $ Z$ respectively.
Show that $ E, F, Z, Y$ are concyclic.
2006 Stanford Mathematics Tournament, 1
A college student is about to break up with her boyfriend, a mathematics major who is apparently more interested in math than her. Frustrated, she cries, ”You mathematicians have no soul! It’s all numbers and equations! What is the root of your incompetence?!” Her boyfriend assumes she means the square root of himself, or the square root of i. What two answers should he give?
2002 Moldova National Olympiad, 2
For every nonnegative integer $ n$ and every real number $ x$ prove the inequality:
$ |\cos x|\plus{}|\cos 2x|\plus{}\ldots\plus{}|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$
1994 Vietnam National Olympiad, 2
$ABC$ is a triangle. Reflect each vertex in the opposite side to get the triangle $A'B'C'$. Find a necessary and sufficient condition on $ABC$ for $A'B'C'$ to be equilateral.
2013 AIME Problems, 15
Let $A,B,C$ be angles of an acute triangle with
\begin{align*}
\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\
\cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9}.
\end{align*}
There are positive integers $p$, $q$, $r$, and $s$ for which \[ \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s}, \] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$.
[i]Note: due to an oversight by the exam-setters, there is no acute triangle satisfying these conditions. You should instead assume $ABC$ is obtuse with $\angle B > 90^{\circ}$.[/i]
1969 IMO, 2
Let $f(x)=\cos(a_1+x)+{1\over2}\cos(a_2+x)+{1\over4}\cos(a_3+x)+\ldots+{1\over2^{n-1}}\cos(a_n+x)$, where $a_i$ are real constants and $x$ is a real variable. If $f(x_1)=f(x_2)=0$, prove that $x_1-x_2$ is a multiple of $\pi$.
1971 Spain Mathematical Olympiad, 4
Prove that in every triangle with sides $a, b, c$ and opposite angles $A, B, C$, is fulfilled (measuring the angles in radians) $$\frac{a A+bB+cC}{a+b+c} \ge \frac{\pi}{3}$$
Hint: Use $a \ge b \ge c \Rightarrow A \ge B \ge C$.
2021 AMC 12/AHSME Fall, 13
The angle bisector of the acute angle formed at the origin by the graphs of the lines $y=x$ and $y=3x$ has equation $y=kx$. What is $k$?
$\textbf{(A)} \: \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)} \: \frac{1+\sqrt{7}}{2} \qquad \textbf{(C)} \: \frac{2+\sqrt{3}}{2} \qquad \textbf{(D)} \: 2\qquad \textbf{(E)} \: \frac{2+\sqrt{5}}{2}$
2004 National Olympiad First Round, 17
Let $R$ and $T$ be points respectively on sides $[BC]$ and $[CD]$ of a square $ABCD$ with side length $6$ such that $|CR|+|RT|+|TC|=12$. What is $\tan (\widehat{RAT})$
$
\textbf{(A)}\ 2\sqrt 3
\qquad\textbf{(B)}\ \sqrt 3
\qquad\textbf{(C)}\ \dfrac 13
\qquad\textbf{(D)}\ \dfrac 12
\qquad\textbf{(E)}\ 1
$
2006 Iran MO (3rd Round), 5
$M$ is midpoint of side $BC$ of triangle $ABC$, and $I$ is incenter of triangle $ABC$, and $T$ is midpoint of arc $BC$, that does not contain $A$. Prove that \[\cos B+\cos C=1\Longleftrightarrow MI=MT\]
2004 Iran MO (2nd round), 1
$ABC$ is a triangle and $\angle A=90^{\circ}$. Let $D$ be the meet point of the interior bisector of $\angle A$ and $BC$. And let $I_a$ be the $A-$excenter of $\triangle ABC$. Prove that:
\[\frac{AD}{DI_a}\leq\sqrt{2}-1.\]
1998 Baltic Way, 11
If $a,b,c$ be the lengths of the sides of a triangle. Let $R$ denote its circumradius. Prove that
\[ R\ge \frac{a^2+b^2}{2\sqrt{2a^2+2b^2-c^2}}\]
When does equality hold?
2006 Princeton University Math Competition, 3
Find the exact value of $\sin 36^o$.
2013 Moldova Team Selection Test, 3
Consider the obtuse-angled triangle $\triangle ABC$ and its side lengths $a,b,c$. Prove that $a^3\cos\angle A +b^3\cos\angle B + c^3\cos\angle C < abc$.
2010 Argentina Team Selection Test, 2
Let $ABC$ be a triangle with $AB = AC$. The incircle touches $BC$, $AC$ and $AB$ at $D$, $E$ and $F$ respectively. Let $P$ be a point on the arc $\overarc{EF}$ that does not contain $D$. Let $Q$ be the second point of intersection of $BP$ and the incircle of $ABC$. The lines $EP$ and $EQ$ meet the line $BC$ at $M$ and $N$, respectively.
Prove that the four points $P, F, B, M$ lie on a circle and $\frac{EM}{EN} = \frac{BF}{BP}$.