1. ## inequality last

ABC is a triangle. BC is perpendicular to CD. |AD|=|DB|=6 units.
|CA|=x units. What is the sum of the biggest and smallest values that x can assume?

I drew paralel AK to DC. |DC|=a and |BC|=s |AK|=2a and |CK|=s

let s=5.9 and a=0.8. Then smallest value of x is 6.
But if we sum 5.9+6 =11.9. Then biggest integer value for x is 11. Then the answer must be 17. But the book says 18.

2. ## Re: inequality last

$s^2+a^2 = 6^2$
$s^2+(2a)^2 = x^2$

Subtracting gives $3a^2 = x^2-6^2$ so $a = \sqrt{\dfrac{x^2-36}{3}}$

Since $a$ is a positive real, it must be that $x^2>36$ so $x>6$.

Next, $s^2 = 6^2-a^2 = 36-\dfrac{x^2-36}{3}$.

$3s^2 = 36\cdot 3 - x^2-36 = 72-x^2$
$s^2 = \dfrac{72-x^2}{3}$
$s = \sqrt{\dfrac{72-x^2}{3}}$

Again, $s$ is a positive real number, so $72>x^2$ implies $6\sqrt{2}>x$.

Now, you have $6<x<6\sqrt{2}$. So, if $x$ is an integer, then $x=7$ or $x=8$.

3. ## Re: inequality last

Isn't it -x^2+36/3? There is - in front of the x^2-36/3.

4. ## Re: inequality last

IF we multiply x^2+s^2=36 with (-),it becomes a^2=x^2-36, Am I wrong?

5. ## Re: inequality last

Originally Posted by kastamonu
Isn't it (-x^2+36)/3? There is - in front of the (x^2-36)/3.
Yes, and when you distribute you get exactly what I showed. I just skipped steps. I also added parentheses because what you wrote was not correct without them.

6. ## Re: inequality last

Originally Posted by kastamonu
IF we multiply x^2+s^2=36 with (-),it becomes a^2=x^2-36, Am I wrong?
Check the math again

7. ## Re: inequality last

Originally Posted by SlipEternal

Since $a$ is a positive real, it must be that $x^2>36$ so $x>6$.

Next, $s^2 = 6^2-a^2 = 36-\dfrac{x^2-36}{3}$.

$3s^2 = 36\cdot 3 - x^2-36 = 72-x^2$

$s^2 = \dfrac{72-x^2}{3}$

$s = \sqrt{\dfrac{72-x^2}{3}}$

Again, $s$ is a positive real number, so $72>x^2$ implies $6\sqrt{2}>x$.

Now, you have $6<x<6\sqrt{2}$. So, if $x$ is an integer, then $x=7$ or $x=8$.
SlipEternal, I was checking some of your math:

$3s^2 = 36\cdot3 - (x^2 - 36) \implies$

$3s^2 = 108 - x^2 + 36 \implies$

$3s^2 = 144 - x^2 \implies$

$s^2 = \dfrac{144 - x^2}{3}$

For s a positive number,

$s = \sqrt{\dfrac{144 - x^2}{3}}$

$144 > x^2 \implies$

$12 > x$

8. ## Re: inequality last

Originally Posted by greg1313
SlipEternal, I was checking some of your math:

$3s^2 = 36\cdot3 - (x^2 - 36) \implies$

$3s^2 = 108 - x^2 + 36 \implies$

$3s^2 = 144 - x^2 \implies$

$s^2 = \dfrac{144 - x^2}{3}$

For s a positive number,

$s = \sqrt{\dfrac{144 - x^2}{3}}$

$144 > x^2 \implies$

$12 > x$
Thanks, that makes sense. I completely misunderstood what the OP was saying. A full description of what he meant along with correct grouping symbols and it is obvious that I made a mistake.

9. ## Re: inequality last

Your solution seems to be very long and time consuming. Follow my steps:
Angle cda is an obtuse angel so c<6 ad+dc<12 then ac<12 7 and 11 is the answer.

10. ## Re: inequality last

Originally Posted by kastamonu
ABC is a triangle. BC is perpendicular to CD. |AD|=|DB|=6 units.
|CA|=x units. What is the sum of the biggest and smallest values that x can assume?

I drew paralel AK to DC. |DC|=a and |BC|=s |AK|=2a and |CK|=s
Suggestion:
make right triangle ABK : BK=6, AK=8, AB=10
so triangle BCD = 3-4-5 and AD=BD=5
AC = sqrt(3^2 + 8^2)

It'll be easier to work with.
Then apply same procedure to your triangle.