# Height of intersection POINT

• Dec 1st 2008, 03:13 AM
mathbuoy
Height of intersection POINT
Two towers (Headbang) with heights a and b meters respectively are 100 meters apart.

What is the height of the intersection point of the line joining the top of each tower to the base of the other?

# show working and explanation PLS
• Dec 1st 2008, 07:15 AM
Soroban
Hello, mathbuoy!

Did you make a sketch?

Quote:

Two towers with heights $\displaystyle a$ and $\displaystyle b$ meters are 100 meters apart.

What is the height of the intersection point of the lines
joining the top of each tower to the base of the other?

Code:

    P *       |*       | *       |  *       |  *       |    *     a |    *    * R       |      *T * |        |      *  |       |    * |  | b       |  *  h| * |       | *    |  *|     Q * - - - * - * S         100-x U x

The towers are: .$\displaystyle PQ = a\,\text{ and }\,RS = b$

The lines cross at $\displaystyle T$, .$\displaystyle h = TU$

Let $\displaystyle US = x\text{, then }QU = 100-x$

Since $\displaystyle \Delta TUS \sim \Delta PQS\!:\;\;\frac{h}{x} = \frac{a}{100} \quad\Rightarrow\quad h \:=\:\frac{ax}{100}$ .[1]

Since $\displaystyle \Delta TUQ \sim \Delta RSQ\!:\;\;\frac{h}{100-x} = \frac{b}{100}\quad\Rightarrow\quad h \:=\:\frac{b(100-x)}{100}$ .[2]

Equate [1] and [2]: .$\displaystyle \frac{ax}{100} \:=\:\frac{b(100-x)}{100} \quad\Rightarrow\quad x \:=\:\frac{100b}{a+b}$

Substitute into [1]: .$\displaystyle h \:=\:\frac{a}{100}\left(\frac{100b}{a+b}\right) \quad\Rightarrow\quad \boxed{h\;=\; \frac{ab}{a+b}}$

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An observation . . .
Divide top and bottom by $\displaystyle ab\!:\;\;h \;=\;\frac{\dfrac{ab}{ab}}{\dfrac{a}{ab} + \dfrac{b}{ab}} \;=\;\frac{1}{\dfrac{1}{a} + \dfrac{1}{b}}$

The answer is one-half the harmonic mean of $\displaystyle a\text{ and }b.$