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no sorry

seriously?

do you know what the word coplanar means?

Let O and H be the circumcenter and orthocenter of triangle ABC, respectively.

Let a, b, and c denote the side lengths as in the picture below:

1.
Find OH if the circumradius is equal to 7 and a^2 + b^2 + c^2 = 432.

\(\text{Let circumradius $R=7$} \\ \text{Let $OA=OB=OC=R$}\)

\(\vec{OH}=\vec{OA}+\vec{OB}+\vec{OC} \quad \text{This is the Sylvester’s relation}\)

\(\begin{array}{|rcll|} \hline \left(\vec{OH}\right)^2 &=& \left(\vec{OA}+\vec{OB}+\vec{OC}\right)^2 \\ OH^2 &=& \underbrace{OA^2+OB^2+OC^2}_{=3R^2} +2\vec{OA}\vec{OB}+2\vec{OA}\vec{OC}+2\vec{OB}\vec{OC} \\ OH^2 &=& 3R^2 +2\vec{OA}\vec{OB}+2\vec{OA}\vec{OC}+2\vec{OB}\vec{OC} \\\\ && 2\vec{OA}\vec{OB} = 2R^2\ \cos(\angle AOB) & c^2 = 2R^2-2R^2\cos(\angle AOB) \\ &&& 2R^2\cos(\angle AOB)=2R^2-c^2 \\ && \mathbf{2\vec{OA}\vec{OB} = 2R^2-c^2} \\\\ && 2\vec{OA}\vec{OC} = 2R^2\ \cos(\angle AOC) & b^2 = 2R^2-2R^2\cos(\angle AOC) \\ &&& 2R^2\cos(\angle AOC)=2R^2-b^2 \\ && \mathbf{2\vec{OA}\vec{OC} = 2R^2-b^2} \\\\ && 2\vec{OB}\vec{OC} = 2R^2\ \cos(\angle BOC) & a^2 = 2R^2-2R^2\cos(\angle BOC) \\ &&& 2R^2\cos(\angle BOC)=2R^2-a^2 \\ && \mathbf{2\vec{OB}\vec{OC} = 2R^2-a^2} \\\\ OH^2 &=& 3R^2 +2R^2-c^2+2R^2-b^2+2R^2-a^2 \\\\ \mathbf{OH^2} &=& \mathbf{9R^2 -\left(a^2+b^2+c^2\right)} \\ \hline OH^2 &=& 9R^2 -\left(a^2+b^2+c^2\right) \quad | \quad R=7,\ a^2 + b^2 + c^2 = 432 \\ OH^2 &=& 9*7^2 -432 \\ OH^2 &=& 9 \\ \mathbf{OH} &=& \mathbf{3} \\ \hline \end{array}\)

also this one answered by one of the millons of geust

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ive found this one and thought it was really good because it also helped me a lot

very good guest and thank you even though im not the one who wrote the question if you were a memeber i would give you a point

edit:

i did any ways

Thank you dgfgrafgdfge111

Thank you dgfgrafgdfge111

In order for a fraction to have a terminating decimal representation the denominator must 

have factors of only powers of 2 or 5.

when n=24, n+101=125 which is 5³

This appears to be the smallest positive integer meeting the criterion.

Box Number

1    2    3    4    5    6

1 = 5^2 - 5 =20

2 = 7^2 - 6 =43

3 = 9^2 - 7 =74

4 =11^2 - 8 = 113

5 =13^2 - 9 = 160

6 =15^2 - 10 = 215

You can also use this "closed form" formula : a_n = 4 n^2 + 3 n - 2 = a(7) =4*7^2  +  3*7  -  2 = 215 

    x = 0.111111.....                 y = 0.01010101......                  z = 0.0001000100010001.....

10x = 1.11111.....             100y = 1.01010101.....         10000z = 1.000100010001.....

  9x = 1                              99y = 1                                9999z = 1

   x = 1/9                                y = 1/99                                   z = 1/9999

   x+y+z = 1213/9999

what are you asking in this question?

I tried this, but I'm still not sure how to do it.

Wow heureka! You have a greatly presented solution!!!

Sum if n terms of a geometric sequence: a1(1-r^n)/(1-r)=168(1-(3/4)^8)/(1/4)=604.724121094.

Melody, school is over. It's the summer!!!! How could I have any homework? These are just problems in a practice book for a math competition that I want to do better in. Even if I had homework that I didn't know how to do, I would ask my teacher for help, not post it here!

I understand this combination, but how did you find this from the beginning?

Thanks for the help Heureka!!!

I never saw that. Thanks!!!

I don't think this is correct

\(0.0001 = 10^{-4}\\ \frac{10^{46}}{10^{-4}}=10^{50}\\ \sqrt{10^{50}}=10^{25}\\10^{-6}\times 10^{25}=10^{19}\\\text{ Hence : }n=19\)

Hint,

There are 12 numbers in the first 'circle'

How many in the second?

How many in the third?

What is the pattern for the 4th etc

I have not worked it out, but that is how I would do it. 

When the greatest common divisor and least common multiple of two integers are multiplied, the product is 180. How many different values could be the greatest common divisor of the two integers?

Let the numbers be ax and ay   where  x and y are relatively prime and a, x, and y are all positive integers.

So the GCD is    \(a\)

and the lowest commone multiple is     axy

The product of these is      \(a^2xy\)

So we have

\(a^2xy=180\\ a^2 \;\;must \;\;be\;\;1,4,9, or \; 36 \;\text{no other square integerwill make this true}\\ \text{So a = 1, -1, 2, -2, 3,- 3, 6 or - 6}\\ \)

So there seems to be 8 possible values for the greatest common divisor.

OSL4#23

\(\large{ \begin{array}{r|r|r|r|r|r|r|r|r|r|} \hline ^{1^\nearrow} & ^{6^\nearrow} & ^{11^\nearrow} & \color{red}- & {\color{red}-}^{20^\nearrow}& ^{42^\nearrow}& ^{92^\nearrow}& ^{192^\nearrow}&^{B=372^\nearrow} \\ \hline ^{1^\nearrow} & ^{5^\nearrow} & {\color{red}-}^{5^\nearrow} & \color{red}- & {\color{red}-}^{9^\nearrow}& ^{22^\nearrow}& ^{50^\nearrow}& ^{100^\nearrow}&^{180^\nearrow} \\ \hline ^{1^\nearrow} & ^{4^\nearrow} & \color{red}- & \color{red}-& {\color{red}-}^{4^\nearrow}& ^{13^\nearrow}& ^{28^\nearrow}& ^{50^\nearrow}&^{80^\nearrow} \\ \hline ^{1^\nearrow} & ^{3^\nearrow} & \color{red}- & \color{red}-& {\color{red}-}^{4^\nearrow}& ^{9\nearrow} & ^{15^\nearrow}& ^{22^\nearrow}&^{30^\nearrow} \\ \hline ^{1^\nearrow} & ^{2^\nearrow} & ^{3^\nearrow} & \color{red}-& {\color{red}-}^{4^\nearrow}& ^{5^\nearrow} & ^{6^\nearrow} & ^{7^\nearrow}& ^{8^\nearrow} \\ \hline ^{A^\nearrow} & ^{1^\nearrow} & ^{1^\nearrow} & & ^{1^\nearrow}& ^{1^\nearrow} & ^{1^\nearrow} & ^{1^\nearrow}& ^{1^\nearrow}\\ \end{array} }\)

We have triangle \(\triangle ABC\) where \(AB = AC\) and \(AD\) is an altitude.
Meanwhile, \(E\) is a point on \(AC\) such that \(AB \parallel DE\).
If \(BC = 12\) and the area of  \(\triangle ABC\) is \(180\), what is the area of \(ABDE\)?

\(\text{Let $AB = AC $ } \\ \text{Let $DB = CD =\dfrac{BC}{2} = 6$ } \)

\(\begin{array}{|rcll|} \hline \text{area of } \triangle ABC =180 &=& \dfrac{BC*AD}{2} \\ 180 &=& \dfrac{12*AD}{2} \\ 180 &=& 6AD \\ \mathbf{ AD } &=& \mathbf{30} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{CD}{ED} &=& \dfrac{BC}{AB} \\\\ \dfrac{6}{ED} &=& \dfrac{12}{AB} \\\\ \dfrac{ED}{6} &=& \dfrac{AB}{12} \\\\ \mathbf{ED }&=& \mathbf{\dfrac{AB}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{area of } ABDE &=& \dfrac{AB+ED}{2}\times H \quad| \quad H=\dfrac{DB*AD}{AB}=\dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \left(\dfrac{AB+ED}{2}\right)\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \left(\dfrac{AB+\dfrac{AB}{2}}{2}\right)\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& \dfrac{3}{4}AB\times \dfrac{6*30}{AB} \\\\ \text{area of } ABDE &=& 3* 3*15 \\\\ \mathbf{\text{area of } ABDE} &=& \mathbf{135} \\ \hline \end{array}\)