2007 tangent circles inscribed in angle (2007 Denmark MO

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0 replies

jlacosta

Apr 2, 2025

0 replies

Inequalities

sqing 5

N 2 minutes ago by sqing

Let $a,b$ be reals such that $a^2+ab+b^2 =3$ . Prove that
$\frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$ $\frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$ $\frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$ $\frac{19}{ 6}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$ Let $a,b$ be reals such that $a^2-ab+b^2 =1$ . Prove that
$\frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$ $\frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$ $\frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$ $\frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$

5 replies

1 viewing

sqing

Wednesday at 4:52 AM

sqing

2 minutes ago

simplfy this

Miranda2829 6

N an hour ago by Miranda2829

4b+13/ -4b+15 = 1/6

anyone can help on the steps to do this ? thank u

6 replies

Miranda2829

Wednesday at 3:09 AM

Miranda2829

an hour ago

Geometry

AlexCenteno2007 3

N 6 hours ago by vanstraelen

Let ABC be an isosceles triangle with AB = AC and M the midpoint of BC. Consider a point E outside the triangle such that BE = BM and CE perpendicular to AB. The point of intersection of the perpendicular bisector of segment EB with the circumcircle of triangle AMB, which is on the same side as A with respect to BE, is point F. Show that angle FME = 90°

3 replies

AlexCenteno2007

Yesterday at 3:49 AM

vanstraelen

6 hours ago

Geometry

AlexCenteno2007 0

Yesterday at 6:28 PM

Let A, B, C, and D be four distinct points on a straight line, in that order. The circles with diameters AC and BD intersect at X and Y. The straight line XY intersects BC at Z. Let P be a point on XY distinct from Z. The straight line CP intersects the circle with diameter AC at C and M, and the straight line BP intersects the circle with diameter BD at B and N. Show that AM, DN, and XY are aligned.

0 replies

AlexCenteno2007

Yesterday at 6:28 PM

0 replies

No more topics!

2007 tangent circles inscribed in angle (2007 Denmark MO - Mohr Contest p4)

parmenides51 2

N Aug 15, 2020 by Hamroldt

The figure shows a $60^o$ angle in which are placed $2007$ numbered circles (only the first three are shown in the figure). The circles are numbered according to size. The circles are tangent to the sides of the angle and to each other as shown. Circle number one has radius $1$ . Determine the radius of circle number $2007$ .
IMAGE

2 replies

parmenides51

Aug 14, 2020

Hamroldt

Aug 15, 2020

2007 tangent circles inscribed in angle (2007 Denmark MO - Mohr Contest p4)

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parmenides51

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parmenides51

#1 h Aug 14, 2020, 9:16 PM

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https://1.bp.blogspot.com/-1bsLIXZpol4/Xzb-Nk6ospI/AAAAAAAAMWk/jrx1zVYKbNELTWlDQ3zL9qc_22b2IJF6QCLcBGAsYHQ/s0/2007%2BMohr%2Bp4.png

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hopetosuccess9102

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hopetosuccess9102

#2 h Aug 14, 2020, 10:31 PM

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i might be wrong

Let Point $A$ be the vertex of the $60^o$ angle, $O_i$ be the center of Circle $i$ , and $T_i$ be the point on Circle $i$ closest to Point $A$ . We notice that if we connect $A$ , $O_i$ , and the point where Circle $i$ is tangent to a line, we get a $30$ - $60$ - $90$ triangle. Consider the triangle with point $O_1$ and let $r_i$ denote the radius of Circle $i$ . We see that the point $T_1$ is on $AO_1$ and we note that $O_1T_1=r_1=1$ . We also see that the length of $O_1$ to the point tangent to Circle $1$ on the line is $r_1=1$ . From the properties of a $30$ - $60$ - $90$ triangle, this length is $\frac 12(AO_1)$ , so $AO_1=2$ . We see that $AO_1=AT_1+O_1T_1=AT_1+1$ and $AT_1=O_1T_1=r_1$ . Continuing in this fashion, we see that $r_2=3$ since $AT_2=3$ . To generalize, we see that $2r_i=AT_i+O_iT_i=AT_i+r_i$ due to the properties of a $30$ - $60$ - $90$ triangle, so $AT_i=r_i$ and $r_{(i+1)}=AT_{(i+1)}=AT_i+T_iT_{(i+1)}=3r_i$ . So, $r_{2007}=3r_{2006}=9r_{2005}=\cdots=3^{2006}(r_1)=\boxed{3^{2006}}$ .

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Hamroldt

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Hamroldt

#3 h Aug 15, 2020, 7:57 AM

Y by

The answer is $3^{2006}$ which is easily proven by induction as well

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