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4

Hint If $P$ is inside $\Delta ABC$, the angles must sum to $360^\circ$, if $P$ is outside, the angles can never be all equal. Can you now prove that $P$ must be unique? (It even has a special name)


3

This is how I interpret the problem: The task is to find the length of the red line.


3

We know $A \hat{D}C = 90 \unicode{xb0}$ and we are given that $\hat{A}= 4 \alpha$. Now, in triangle $ADC$, we require $$ 4\alpha + \alpha + 90\unicode{xb0}= 180 \unicode{xb0} \\ \therefore 5\alpha + 90 \unicode{xb0} = 180 \unicode{xb0}$$ All that is left now, is to solve for $\alpha$ \begin{align} 5\alpha &= 180 \unicode{xb0} - 90 \unicode{xb0}\\ ...


3

Writing $\cos 2x = 1 -\sin^2 x$ and using the sine rule $\frac{a}{\sin \alpha} = 2R$, the inequality turns out to be equivalent to $$ a^2 + b^2 + c^2 \leq 9R^2 $$ where $a,b,c$ are the sides of the triangle and $R$ is the circumradius. This inequality is sometimes known as Leibniz's inequality. For this inequality, a geometric proof is possible. Let $O$ ...


3

Here's an approach using multiple applications of an Extended Ceva's Theorem that I introduced in a previous answer. The theorem gives a condition for concurrence of (almost) any three lines passing through the edges of a triangle, not just those through vertices. Extended Ceva's Theorem. Consider points $D_1$, $D_2$, $E_1$, $E_2$, $F_1$, $F_2$ on the ...


3

Hint: Try to prove that $\Delta ABD \cong \Delta ACM$ using parts $(1),(2)$ of the problem.


3

we will use that an arc makes same angle on the circumference. $\angle BMC = \angle BAC = 60^\circ$ angle made by arc $BC$ similarly $\angle BMA = \angle BCC = 60^\circ$ therefore $\angle BMC = \angle BMA$ and that shows $BM$ bisects $\angle AMC$ (2) look at the isoscles triangle $AMD$ with one base angle $AMD = 60^\circ$ that makes $AMD$ equilateral ...


3

$(3)$ is a consequence of Ptolemy's theorem, from which: $$ MB\cdot AC = MA\cdot BC + MC\cdot AB. $$ Since $AC=BC=AB$, $$ MB = MA+MC $$ follows.


2

When dealing with literal triangles in the plane, the inequality is strict so as to prevent degenerate triangles. To get a magnitude of $7$, draw $\textbf{b}$ so that it points east and draw $\textbf{a}$ parallel (so that it also points east). Then their difference $\textbf{b} - \textbf{a}$ is a vector of magnitude $7$ that points east. Note that the vectors ...


2

Let $\angle PBM= \alpha$ and $\angle NCQ=\beta$. Then, as $BP$ is the median of $\triangle ABM$, we have that $\cot \alpha +\cot A=\cot B+\cot (\pi-A)$, which gives $\cot \alpha = 2\cot A+\cot B$ and similarly $\cot \beta = 2\cot A+\cot C$. For convenience, let us take $\cot A=c_1$, $\cot B=c_2$ and $\cot C=c_3$. By the addition formula we have: ...


2

Here is a complete proof along the lines of my comments to the OP. Most of it is well-known stuff, but it probably doesn't hurt to build it from scratch. First of all, the problem follows from the following claim: Theorem 1. Let $A$, $B$, $C$ and $D$ be four points on a circle. Let $X=AB\cap CD$ and $Y=BC\cap DA$. (I use the notation $\ell\cap m$ to denote ...


2

By the angle bisector theorem, we have: $$\frac{CE}{AF}=\frac{CD\cos C}{AD\cos A}=\frac{BC\cos C}{AB\cos A}=\frac{CI}{AI}$$ As $BF=BE$, by the converse of Ceva's theorem, we have that $AE, CF$ and $BI$ are concurrent.


2

Let $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AB}$, respectively, noting that $N$ is also the intersection of the lateral edges of trapezoid $\square BCGE$. Now, if $\angle AEC\cong \angle DGC$, then $\angle BEC\cong\angle BGC$. This implies that $\square BCGE$ is cyclic; since it's a trapezoid, it's therefore necessarily an isosceles ...


2

By Holder $\left(\sum\limits_{cyc}\frac{\sqrt{b+c-a}}{a}\right)^2\sum\limits_{cyc}a^2(b+c-a)^2\geq(a+b+c)^3$. Hence, it remains to prove that $abc(a+b+c)\geq\sum\limits_{cyc}a^2(b+c-a)^2$, which is $\sum\limits_{cyc}(a-b)^2(a+c-b)(b+c-a)\geq0$, which is obvious.


2

Where is it located? Unfortunately I couldn't find a closed formula for the trilinears yet. Here is what I have so far. Let the outer triangle have edge lengths proportional to $a_1:b_1:c_1$. The trilinear coordinates of the recursive contact center with respect to that triangle would be called $x_1:y_1:z_1$. Now the edge lengths of the contact triangle ...


2

Let $A=(x_1, y_1)$, $B=(x_2, y_2)$ and $C=(x_3, y_3)$ are the vertices of a triangle $ABC,$ $c,$ $a$ and $b$ are the lengths of the sides $AB,$ $BC$ and $AC$ respectively. Then coordinates of center of ex-circle opposite to vertex $A$ are given as $$I_1(x, y) =\left(\frac{–ax_1+bx_2+cx_3}{–a+b+c},\frac{–ay_1+by_2+cy_3}{–a+b+c}\right).$$ Similarly ...


2

Perhaps you'll find it easier to compute the area by expressing it as half the magnitude of the vector product $\left\langle b,a,0\right\rangle \times \left\langle x,y,0\right\rangle$. Indeed, if $\theta$ is the interior angle between the base and the side $\ell$ from the origin to $(x,y)$, then $h = \ell\sin \theta$. So the area is $(1/2)r\ell\sin \theta$. ...


1

If we denote the point $(x,y)$ by $(x_1,y_1)$ to avoid confusion, we can take the base of the triangle to be $\;\;\;B=\sqrt{x_1^2+y_1^2}$. The line through $(0,0)$ and $(x_1,y_1)$ has equation $y=\frac{y_1}{x_1}x$ (assuming $x_1\ne0$), so the perpendicular line through $(b,a)$ has equation $y-a=-\frac{x_1}{y_1}(x-b)$ or $y=-\frac{x_1}{y_1}(x-b)+a$ (assuming ...


1

This is an easy problem to solve with trilinear coordinates. We have $I=[1,1,1]$, $D=\left[\frac{1}{a},\frac{1}{b},0\right]$ and $E=\left[\frac{1}{a},0,\frac{1}{c}\right]$. Moreover, we have $H=[1,0,\mu]$ and $G=[1,\eta,0]$, with $\det(D,I,H)=\det(E,I,G)=0$, so: $$ H=\left[\frac{1}{b}-\frac{1}{a},0,\frac{1}{b}\right],\qquad ...


1

My belief is that the statement is false. Assume it is true, that OIH is a straight line. Then, AOD and IOH are perpendicular, meaning that AD is perpendicular to IH. Similarly, this implies that BE and CF (where E, F are the foot of the perpendiculars from I to AC, and AB respectively) are perpendicular to IH. But this is clearly not possible. Hence, ...


1

By using trilinear coordinates (today is the day) we have that a point belongs to the Euler line $OH$ iff its trilinear coordinates $[x:y:z]$ satisfy $$ \sum_{cyc} \sin(2A)\sin(B-C)x = 0. \tag{1}$$ For the incenter $I=[1:1:1]$, the last identity is equivalent to: $$ \left(\sin(A)+\sin(B)+\sin(C)\right)\prod_{cyc}\sin\frac{A-B}{2},\tag{2} $$ so, if we assume ...


1

Consider the diagram: $d=\overline{CE}=\overline{CF}$. Note that $c=\overline{AB}=(d-a)+(d-b)$. Therefore, $$ d=\frac{a+b+c}2\tag{1} $$ Furthermore, $d=\overline{CD}\cos(\theta/2)$ and $\overline{CH}=d\cos(\theta/2)$; therefore, $\overline{CH}=\overline{CD}\cos^2(\theta/2)$. The Law of Cosines gives $$ \cos(\theta)=\frac{a^2+b^2-c^2}{2ab}\tag{2} $$ so ...


1

Consider the following diagram: Using Barycentric Coordinates, we get that the coordinates of $D$ to be $$ D=\frac{|\triangle BCD|A+|\triangle ACD|B-|\triangle ABD|C}{|\triangle BCD|+|\triangle ACD|-|\triangle ABD|}\tag{1} $$ The weight from $\triangle ABD$ is negative since that triangle is outside of $\triangle ABC$. Since each of the triangles in ...


1

It seems the following. Let the triangle with vertices $z_1, z_2, z_3$ has sides $a, b, c$, area $S$ and radius $R$ of the circumcircle. Then $R=\frac {abc}{4S}$. The formula for oriented area $$2S=\left|\begin{array}{} x_1-x_3 & x_2-x_3\\ y_1-y_3 & y_2-y_3\\ \end{array}\right|, $$ yields $4S=2\Im(\bar z_1z_2+\bar z_2z_3+\bar z_3z_1)$. So $R=1$ ...


1

I'm assuming that the goal here is to show All five points $I$, $B$, $C$, $D^\prime$ $E^\prime$ are concyclic if and only if $|\overline{AB}| + |\overline{AC}| = 3|\overline{BC}|$. (If we were only to show that $I$, $B$, $C$, $D^\prime$ are concyclic, then, by symmetry of argument, $E^\prime$ would have to lie on the common circle, and vice-versa. ...


1

I'll get you started: For (2), we use $\text{Area} = \frac{1}{2}ab\sin(C)$, where $C$ is the angle between the sides with length $a$ and $b$. So, $$ [CDE] = \frac{1}{2} \times 3 \times 5 \times \sin(\angle ACB) = 7, $$ Use this to find $\sin(\angle ACB)$, from which you can use the same formula to find $[ABC]$. For (3), the same formula can get you $\angle ...


1

Rotation of a shape $S$ which has an area $A$ can result in different volumes depending on the shape of $S$, even if the area of $S$ stays constant. As an example, consider a rectangle with sides of length $1$ and $2$. If you rotate it around the shorter side, you get a cylinder of volume $\pi r^2 h = \pi(2)^2(1) = 4\pi$. But, if you rotate it around the ...


1

Combine $$cos(2a)+cos(2b)+cos(2c)=-4cos(a)cos(b)cos(c)-1$$ and $$cos(a)cos(b)cos(c) \leq \frac{1}{8}.$$ Both formulas can be derived by using elementary methods. For the first formula you only the addition formulas for cosine. Similarly, for the inequality.


1

This is basically Langley's Adventitious Angles problem, a problem that first appeared in a $1922$ Mathematical Gazette, and is well known for being extremely difficult despite its simplistic appearance. Here is a nice piece by piece solution with diagrams. For the sake of completeness, I will go ahead and add the Wikipedia solution which is attributed to ...


1

Consider that: $$\frac {AF}{BF}=\frac{FA}{FD}\cdot \frac{FD}{FB}=\cot A\cdot \tan \frac{B}{2}$$ $$\frac {BE}{EC}=\frac{BE}{DE}\cdot \frac{DE}{CE}=\cot \frac{B}{2}\cdot \tan C$$ $$\frac{IC}{IA}=\frac{IC}{IB}\cdot \frac{IB}{IA}=\cot C\cdot \tan A$$ Multiply those equalities side by side, $AE,BF,CI$ are concurrent by Ceva theorem.



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