Geo Mock #1

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0 replies

jlacosta

Wednesday at 3:18 PM

0 replies

New geometry problem

titaniumfalcon 2

N 8 minutes ago by fruitmonster97

Post any solutions you have, with explanation or proof if possible, good luck!

2 replies

titaniumfalcon

Yesterday at 10:40 PM

fruitmonster97

8 minutes ago

.problem.

Cobedangiu 2

N 8 minutes ago by Lankou

Find the integer coefficients after expanding Newton's binomial:
$(\frac{3}{2}-\frac{2}{3}x^2)^n (n \in Z)$

2 replies

Cobedangiu

6 hours ago

Lankou

8 minutes ago

Inequalities

sqing 23

N 31 minutes ago by sqing

Let $a,b,c\geq 0$ and $a+b+c=1$ . Prove that
$a^2+b^2+ ab +21abc\leq\frac{512}{441}$ Equality holds when $a=b=\frac{38}{21},c=\frac{5}{214}.$
$a^2+b^2+ ab +19abc\leq\frac{10648}{9747}$ Equality holds when $a=b=\frac{22}{57},c=\frac{13}{57}.$
$a^2+b^2+ ab +22abc\leq\frac{15625}{13068}$ Equality holds when $a=b=\frac{25}{66},c=\frac{8}{33}.$

23 replies

sqing

Mar 26, 2025

sqing

31 minutes ago

Inequalities

sqing 3

N an hour ago by sqing

Let $a, b,c\geq 0$ and $2a+3b+ 4c=11.$ Prove that
$a+ab+abc\leq\frac{49}{6}$ Let $a, b,c\geq 0$ and $2a+3b+ 4c=10.$ Prove that
$a+ab+abc\leq\frac{169}{24}$ Let $a, b,c\geq 0$ and $2a+3b+ 4c=14.$ Prove that
$a+ab+abc\leq\frac{63+5\sqrt 5}{6}$ Let $a, b,c\geq 0$ and $2a+3b+ 4c=32.$ Prove that
$a+ab+abc\leq48+\frac{64\sqrt{2}}{3}$

3 replies

sqing

Apr 1, 2025

sqing

an hour ago

No more topics!

Geo Mock #1

Bluesoul 2

N Yesterday at 4:13 AM by jb2015007

Consider the rectangle $ABCD$ with $AB=4$ . Point $E$ lies inside the rectangle such that $\triangle{ABE}$ is equilateral. Given that $C,E$ and the midpoint of $AD$ are on the same line, compute the length of $BC$ .

2 replies

Bluesoul

Apr 1, 2025

jb2015007

Yesterday at 4:13 AM

Geo Mock #1

G H J

Reveal topic tags

geometry

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Bluesoul

890 posts

Bluesoul

#1 h Apr 1, 2025, 6:58 AM

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Sedro

5819 posts

Sedro

#2 h Apr 1, 2025, 4:30 PM

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Solution

Let the midpoint of $AD$ be $F$ . Set $A=(0,4)$ , $B=(0,0)$ , and $C=(2k,0)$ for some positive real $k$ . Then, $E$ has coordinates $(2\sqrt{3},2)$ and $F$ has coordinates $(k,4)$ . The line passing through $E$ and $F$ has equation $y = -\tfrac{4}{k}(x-2k)$ . Plugging in $(x,y)=(2\sqrt{3},2)$ , we find that $BC = 2k= \boxed{\tfrac{8\sqrt{3}}{3}}$ .

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jb2015007

1777 posts

jb2015007

#3 h Yesterday at 4:13 AM

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Solution (Basic Geo)

Assign $F$ to be the midpoint of $AD$ and $X$ to be the midpoint of $AB$ . Now, we let $x$ be the perpendicular distance from $E$ to $CD$ . Dropping this altitude gives $2$ similar triangles in a $1:2$ ratio, so $FD = 2x$ , therefore $AD = 4x$ . Now, we can try to $BC$ again in terms of $x$ , in a different way. We have, by $30-60-90$ right triangles that, $XE = \dfrac{4}{2} \cdot \sqrt3 = 2\sqrt3$ . Now, we can write an equation! We have: $4x = x+2\sqrt3 \Longrightarrow 3x = 2\sqrt3 \Longrightarrow 4x = \dfrac{4}{3} \cdot 2\sqrt3 = \boxed{\dfrac{8\sqrt{3}}{3}}$

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