Geo Mock #5

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0 replies

jlacosta

Apr 2, 2025

0 replies

Inequalities

sqing 2

N 28 minutes ago by DAVROS

Let $a,b,c\in [0,1]$ . Prove that
$(a+b+c)\left(\frac{1}{a^2+3}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\right)\leq \frac{19}{4}$ $(a+b+c)\left(\frac{1}{a^2+ 4}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\right)\leq \frac{23}{5}$ $(a+b+c)\left(\frac{1}{a^2+ \frac{5}{2}}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\right)\leq \frac{34}{7}$ $(a+b+c)\left(\frac{1}{a^2+ \frac{7}{2}}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\right)\leq \frac{14}{3}$

2 replies

sqing

Yesterday at 3:33 AM

DAVROS

28 minutes ago

Stylish Numbers

pedronis 0

29 minutes ago

A positive even integer $n$ is called stylish if the set $\{1, 2, \ldots, n\}$ can be partitioned into $\frac{n}{2}$ pairs such that the sum of the elements in each pair is a power of $3$ . For example, $6$ is stylish because the set $\{1, 2, 3, 4, 5, 6\}$ can be partitioned as $\{1,2\}, \{3,6\}, \{4,5\}$ , with sums $3$ , $9$ , and $9$ respectively. Determine the number of stylish numbers less than $3^{2025}$ .

0 replies

pedronis

29 minutes ago

0 replies

New geometry problem

titaniumfalcon 4

N an hour ago by titaniumfalcon

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

4 replies

titaniumfalcon

Apr 3, 2025

titaniumfalcon

an hour ago

A rather difficult question

BeautifulMath0926 1

N 2 hours ago by jasperE3

I got a difficult equation for users to solve:
Find all functions f: R to R, so that to all real numbers x and y,
1+f(x)f(y)=f(x+y)+f(xy)+xy(x+y-2) holds.

1 reply

BeautifulMath0926

Today at 11:10 AM

jasperE3

2 hours ago

No more topics!

Geo Mock #5

Bluesoul 3

N Apr 4, 2025 by rhydon516

Consider triangle $ABC$ with $AB=13, BC=14, AC=15$ . Denote the orthocenter of $\triangle{ABC}$ as $H$ , the intersection of $(BHC)$ and $AC$ as $P\neq C$ . Compute the length of $AP$ .

3 replies

Bluesoul

Apr 1, 2025

rhydon516

Apr 4, 2025

Geo Mock #5

G H J

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Bluesoul

892 posts

Bluesoul

#1 h Apr 1, 2025, 7:03 AM

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Consider triangle $ABC$ with $AB=13, BC=14, AC=15$ . Denote the orthocenter of $\triangle{ABC}$ as $H$ , the intersection of $(BHC)$ and $AC$ as $P\neq C$ . Compute the length of $AP$ .

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Sedro

5823 posts

Sedro

#2 h Apr 2, 2025, 3:07 PM

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Solution

Let $D$ be the foot of the altitude from $A$ to $BC$ , and let $A'$ be the reflection of $A$ over $BC$ . Computation gives that $BD=5$ , $CD=9$ , and $AD=A'D=12$ . By orthocenter properties, $\triangle BHD \sim \triangle ACD \sim \triangle A'CD$ . Thus, $DH = \tfrac{(BD)(CD)}{A'D} = \tfrac{15}{4}$ and $AH = 12-DH = \tfrac{33}{4}$ . Now, let $x=AP$ . By Power of a Point at $A$ with respect to $(BHC)$ , we have $15x = (\tfrac{33}{4})(24)$ , and solving yields $x = \boxed{\tfrac{66}{5}}$ .

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fruitmonster97

2459 posts

fruitmonster97

#3 h Apr 3, 2025, 9:03 PM

Y by

xonk!

Solution

Consider the second intersection of $AH$ and $(BHC),$ say $A'.$ Then $A'$ is the reflection of $A$ over $BC.$ Then it's just computation - find $AH,$ PoP, set eq and win.

Let $X\in BC$ s.t. $AX\perp BC,$ and let $Y\in AB$ s.t. $CY\perp AB.$ Then $CY=168/13,$ so $AY^2=15^2-(168/13)^2=(195^2-168^2)/13^2=(363*27)/13^2=99^2/13^2,$ and thus $AY=99/13.$ Then $AXB\sim AYH,$ so $AH=AB\cdot\tfrac{AY}{AX}=\tfrac{99}{12}.$ The power of $A$ wrt $(BHC)$ is thus $AH\cdot AA'=\tfrac{99}{12}\cdot2AX=198.$ The power of $A$ wrt $(BHC)$ is also $AC\cdot AP,$ so $AP=\tfrac{198}{15}=\boxed{\tfrac{66}{5}}$

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rhydon516

549 posts

rhydon516

#4 h Apr 4, 2025, 7:01 AM • 1 Y

Y by Bluesoul

clean

A quick angle chase gives $\angle APB=\angle A$ , and the rest is computation by whatever method you choose.

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