2015 Mathcounts State Prep : Inscribed Cricle Radius and Similar Triangles

Question : $\mathrm{\Delta }$ ABC is an equilateral triangle. Circle "O" is the inscribed circle and it's radius is 15. 

What is the length of the radius of the smaller circle p which is tangent to circle "O" and the two sides?








Here is the link to the basics of inscribed circle radius as well as circumscribed circle radius of an equilateral triangle.

Solution I :
The length of the radius of an inscribed circle of an equilateral triangle is ${\displaystyle \frac{1}{3}}$ of the height so you know AO is ${\displaystyle \frac{2}{3}}$ of the height or 30 (the height is 15 + 30 = 45 unit long)

$\mathrm{\Delta }$ AEP is similar to $\mathrm{\Delta }$ AFO $\to$ ${\displaystyle \frac{r}{15}}={\displaystyle \frac{AP}{30}}$
so $\overline{AP}$ = 2r.
$\overline{AP}+\overline{PO}=30$  $\to$2r + r + 15 = 30   $\to$  3r = 15 so r = 5 
or ${\displaystyle \frac{1}{3}}$ of the larger radius 

Solution II:
$\mathrm{\Delta }$ APE is a 30-60-90 right triangle, so $\overline{AP}$ = 2r
$\overline{PO}$ = r + 15
$\overline{AP}+\overline{PO}$  $\to$  2r + r + 15 = 30  $\to$  3r  =  15 so  r  = 5 
or ${\displaystyle \frac{1}{3}}$ of the larger radius 

This is an AMC-10 question.

$\mathrm{\Delta }$ ABC is an isosceles triangle.

The radius of the smaller circle is 1 and the radius of the larger circle is 2,

A: what is the length of $\overline{AP}$ ?

B. what is the area of $\mathrm{\Delta }$ ABC?







Solution for question A: 
$\mathrm{\Delta }$ AEP is similar to $\mathrm{\Delta }$ AFO $\to$ ${\displaystyle \frac{1}{2}}={\displaystyle \frac{AP}{AP+3}}$
 2$\overline{AP}$ =  $\overline{AP}$ + 3 $\to$ AP = 3

Using Pythagorean theorem, you can get $\overline{AE}$ = $2\sqrt{2}$
$\mathrm{\Delta }$ AEP is similar to $\mathrm{\Delta }$ ADC [This part is tricky. Make sure you see that !!]
 $\to$ ${\displaystyle \frac{1}{\overline{DC}}}={\displaystyle \frac{AE}{AD}}$ = ${\displaystyle \frac{2\sqrt{2}}{8}}$
$\overline{DC}$ = $2\sqrt{2}$  and $\overline{BC}$ = 2 * $2\sqrt{2}$ = $4\sqrt{2}$
The area of  $\mathrm{\Delta }$ ABC =  ${\displaystyle \frac{1}{2}}$*$4\sqrt{2}$ * 8 =  $16\sqrt{2}$

Question: If you know the length of x and y, and the whole length of $\overline{AB}$,

A: what is the ratio of a to b and 

B: what is the length of z.







Solution for question A:
$\mathrm{\Delta }$ABC and $\mathrm{\Delta }$AFE are similar so ${\displaystyle \frac{z}{x}}={\displaystyle \frac{b}{a+b}}$. -- equation 1
Cross multiply and you have z ( a + b ) = bx

$\mathrm{\Delta }$BAD and $\mathrm{\Delta }$BFE are similar so ${\displaystyle \frac{z}{y}}={\displaystyle \frac{a}{a+b}}$. -- equation 2
 Cross multiply and you have z ( a + b ) = ay

bx = ay so ${\displaystyle \frac{x}{y}}={\displaystyle \frac{a}{b}}$  same ratio

Solution for question B: 
Continue with the previous two equations, if you add equation 1 and equation 2, you have:
${\displaystyle \frac{z}{x}}+{\displaystyle \frac{z}{y}}={\displaystyle \frac{b}{a+b}}+{\displaystyle \frac{a}{a+b}}$
${\displaystyle \frac{zy+zx}{xy}}=1$ $\to$ z = ${\displaystyle \frac{xy}{x+y}}$

Applicable question: 

$\overline{CD}=15$ and you know $\overline{DB}:\overline{BC}=20:30=2:3$ 
 so  $\overline{DB}=6$ and $\overline{BC}=9$ 

$\overline{AB}={\displaystyle \frac{20\times 30}{(20+30)}}$ = 12