2
$\begingroup$

The incircle of trianlge ABC touches the sides AB, AC and BC at points P, N, and M respectively. Denote AP=AN=x, BM=BP=y and CM=CN=z. Segment UV is tangent to the incircle at point X and parallel to the side AC.

Prove $\cfrac{UV}{AC}$= $\cfrac{y}{x+y+z}$ enter image description here

What I have so far $\triangle$ ABC $\sim$ $\triangle$ BUV which has gotten $\cfrac{BV}{BC}$ = $\cfrac{BU}{AB}$ =$\cfrac{UV}{AC}$

any help would be appreciated!

$\endgroup$
  • $\begingroup$ Do you mean $AP = AN = x$? $\endgroup$ – gt6989b May 19 '17 at 19:09
  • $\begingroup$ @gt6989b Yes! sorry! $\endgroup$ – Parley May 19 '17 at 19:10
1
$\begingroup$

Let $a$, $b$, $c$ be the sidelengths of $BC$, $CA$, $AB$, let $h_1$ be the height from $B$ to $AC$, and let $h_2$ be the height from $B$ to $UV$. Note that $\dfrac{UV}{AC} = \dfrac{h_2}{h_1}$ because of the similar triangles you mentioned. Also, $h_2 = h_1 - 2r$ where $r$ is the inradius of triangle $ABC$. Hence $$\dfrac{UV}{AC} = \dfrac{h_1-2r}{h_1} = 1-\dfrac{2r}{h_1}.$$ Now let $K$ be the area of triangle $ABC$. We know $2K = h_1b = r(a+b+c)$, so $$\dfrac{r}{h_1} = \dfrac{2K/h_1}{2K/r} = \dfrac{b}{a+b+c}.$$ Thus $$\dfrac{UV}{AC} = 1-\dfrac{2b}{a+b+c} = \dfrac{a-b+c}{a+b+c} = \dfrac{y}{x+y+z}.\ \blacksquare$$ (Notes: The last step follows from $x = (b+c-a)/2, y=(c+a-b)/2, z=(a+b-c)/2$, which is well-known. To get the formula $2K = r(a+b+c)$, note that $K$ is the sum of the areas of triangles $IAB$, $IBC$, and $ICA$.)

$\endgroup$
  • $\begingroup$ how did u get $\cfrac{UV}{AC}$=$\cfrac{h_{2}}{h_{1}}$ $\endgroup$ – Parley May 19 '17 at 21:23
  • $\begingroup$ Also, where did the whole "notes" section come from? what is x, y and z? $\endgroup$ – Parley May 19 '17 at 21:37
  • $\begingroup$ $UV/AC = h_2/h_1$ comes from the similar triangles $ABC$ and $UBV$. $x$, $y$, and $z$ are the same as they are in the diagram in your question. Hope that helps! $\endgroup$ – fractal1729 May 20 '17 at 0:33
  • $\begingroup$ Yes! can u just explain how u arrived at x y and z $\endgroup$ – Parley May 20 '17 at 0:42
  • $\begingroup$ $y+z = a$, $z+x = b$, $x+y = c$ is a system of three equations in the three variables $x,y,z$. Solving the system in terms of $a,b,c$ yields the three formulas for $x,y,z$ that I mentioned in my post. $\endgroup$ – fractal1729 May 20 '17 at 0:46
0
$\begingroup$

\begin{align*} &\text{For triangle ABC}\\[4pt] &\;\;\;{\small{\bullet}}\;\;\text{Let $r$ be the length of the inradius.}\\[4pt] &\;\;\;{\small{\bullet}}\;\;\text{Let $p$ be the perimeter.}\\[4pt] &\;\;\;{\small{\bullet}}\;\;\text{Let $k$ be the area.}\\[4pt] &\;\;\;{\small{\bullet}}\;\;\text{Let $h$ be the length of the altitude from B.}\\[12pt] &\text{Then}\;\;\frac{1}{2}pr = k =\frac{1}{2}|\text{AC}|h\\[4pt] &\implies\;pr = |\text{AC}|h\\[4pt] &\implies\;2(x+y+z)r = (x + z)h\\[4pt] &\implies\;\frac{r}{h} = \frac{x+z}{2(x+y+z)}\\[12pt] &\text{For triangle UBV, Let $h_1$ be the length of the altitude from B.}\\[12pt] &{\text{Then}\;\;h1}= h - |\text{XN}|\\[4pt] &{\phantom{\text{Then}\;\;h1}} = h - 2r\\[12pt] &\text{Then since triangle UBV is similar to triangle ABC,}\\[12pt] &{\phantom{\implies\;}}\frac{|\text{UV}|}{|\text{AC}|} = \frac{h_1}{h}\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = \frac{h-2r}{h}\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = 1 - 2\left(\frac{r}{h}\right)\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = 1 - 2\left( \frac{x+z}{2(x+y+z)}\right)\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = 1 - \frac{x+z}{x+y+z}\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = \frac{(x+y+z) - (x+z)}{x+y+z}\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = \frac{y}{x+y+z}\\[4pt] \end{align*}

$\endgroup$
  • $\begingroup$ where are you drawing the inradius from? and where did the equation $\cfrac{1}{2}pr$= $\cfrac{1}{2}$ $|{AC}|h$ come from? $\endgroup$ – Parley May 19 '17 at 20:21
  • $\begingroup$ @Parley: The formula $$k = \frac{1}{2}pr$$ is a standard formula, and easily proved. And the formula $$k =\frac{1}{2}|\text{AC}|h$$ is just $$k =\frac{1}{2}bh$$ $\endgroup$ – quasi May 19 '17 at 20:27
  • $\begingroup$ @Parley As I noted in my answer, to prove $k = pr/2$, consider the areas of triangle $IAB$, $IBC$, $ICA$ and their sum, $k$. Hope that helps! $\endgroup$ – fractal1729 May 19 '17 at 20:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.