computational with angles 45^o,75^o, AB=CD=1 (Kharkiv City VIII 2018

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0 replies

jlacosta

Apr 2, 2025

0 replies

Inequalities

sqing 4

N an hour 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
$\frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$ $\frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$ $\frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$ $\frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$

4 replies

sqing

Yesterday at 4:52 AM

sqing

an hour ago

Geometry

AlexCenteno2007 1

N 2 hours ago by sunken rock

Let C be a point on a semicircle of diameter AB, and let D be the midpoint of arc AC. Let E and F be the intersections of the altitudes of triangles ABC and DBC with segment BC, respectively. Prove that line BF bisects segment DE.

1 reply

AlexCenteno2007

6 hours ago

sunken rock

2 hours ago

Geometry

AlexCenteno2007 1

N 3 hours ago by sunken rock

In a triangle ABC, the bisector from vertex A intersects side BC at point D. Points M and N are chosen on sides AB and AC, respectively, so that angle ADM is equal to angle ABC, and angle ADN is equal to angle ACB. Line MN intersects AD at point P. Show that the cube of the length of AD is equal to the product of the lengths AB, AP, and AC.

1 reply

AlexCenteno2007

6 hours ago

sunken rock

3 hours ago

purple comet math competition question

AVY2024 5

N 4 hours ago by ehz2701

Given that (1 + tan 1)(1 + tan 2). . .(1 + tan 45) = 2n, find n

5 replies

AVY2024

Tuesday at 11:00 AM

ehz2701

4 hours ago

No more topics!

computational with angles 45^o,75^o, AB=CD=1 (Kharkiv City VIII 2018 - Ukr)

parmenides51 2

N Mar 24, 2020 by sunken rock

Let $ABCD$ be a convex quadrilateral with $\angle A = 45^o, \angle ADC = \angle ACD = 75^o$ and $AB = CD = 1$ . Find the length of the segment $BC$ . Justify the answer.

2 replies

parmenides51

Mar 21, 2020

sunken rock

Mar 24, 2020

computational with angles 45^o,75^o, AB=CD=1 (Kharkiv City VIII 2018 - Ukr)

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parmenides51

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parmenides51

#1 h Mar 21, 2020, 6:16 AM

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Let $ABCD$ be a convex quadrilateral with $\angle A = 45^o, \angle ADC = \angle ACD = 75^o$ and $AB = CD = 1$ . Find the length of the segment $BC$ . Justify the answer.

This post has been edited 1 time. Last edited by parmenides51, Mar 21, 2020, 6:20 AM

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vanstraelen

8965 posts

vanstraelen

#2 h Mar 22, 2020, 10:15 AM

Y by

$\triangle ACD$ is isosceles, so $AD=AC=\frac{1}{2\sin 15^{\circ}}$ .

Rule of cosines in $\triangle ABC\ :\ BC^{2}=1 + \frac{1}{4\sin^{2}15^{\circ}}-2 \cdot 1 \cdot \frac{1}{2\sin 15^{\circ}} \cdot \cos 15^{\circ}$ .

$BC^{2}=1 + \frac{1-4\sin 15^{\circ}\cos 15^{\circ}}{4\sin^{2}15^{\circ}}=1 + \frac{1-2\sin 30^{\circ}}{4\sin^{2}15^{\circ}}=1$ .

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sunken rock

4383 posts

sunken rock

#3 h Mar 24, 2020, 1:03 PM • 1 Y

Y by Mango247

Let $O$ be the circumcenter of $\triangle ACD$ ; notice it is the symmetrical of $B$ across $AC$ , thus $BC=CO=AO=CD=AB=1$ .

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sunken rock

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