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We have a right triangle triangle ABC where the legs AB and BC have lengths 6 and 3sqrt3, respectively. Medians AM and CN meet at point P. What is the length of CP?

MIRB15  Jul 12, 2017
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#1
+75017
+1

Let AM lie on a line

The slope is -6/(1.5 sqrt (3))  =  -4/sqrt (3)

And the equation of this line   y =     -4x/sqrt (3)  + 6

Let CN lie on another line

Ths slope is -3/ 3sqrt (3)  =  -1/sqrt(3)

And the equation of this line is y =  -1x/sqrt(3) + 3

Setting these two lines equal and solving for x will give us the x coordinate for P

-4x/sqrt (3)  + 6 = -1x/sqrt(3) + 3  rearrange

3  =  [ 4 - 1] x / sqrt (3)

3  =  3 x/ sqrt (3)       multiply both sides by  sqrt (3) / 3

sqrt (3)  = x

Subbing this into the second equation to find the y coordinate of P we have that

y =   -1 sqrt (3)  /sqrt(3) + 3 =  2

So......using the distance formula, CP =

sqrt [ (3sqrt(3) - sqrt (3))^2  + (2 - 0)^2 ]  = sqrt [ (2sqrt(3))^2  + (2)^2 ] =

sqrt [ 12 + 4]  =

sqrt [16]  =   4

Here's a pic

CPhill  Jul 13, 2017
edited by CPhill  Jul 13, 2017
#2
+75017
+1

Here's another method to solve this......

Connect  MN......Since, this segment divides the sides AB and BC of triangle ABC proportionally, then MN is parallel to AC

And AM  is a transversal  between two parallel lines so that angles MAC and AMN  form equal alternate interior angles.......and angle APC = angle MPN since they are both vertical angles

So......by angle-angle congruency, triangle APC is similar to triangle MPN

Also, since AB and MN are parallel, then angle NMB = angle ACB  and  angle NBM = angle ABC.....so, again, by AA congruency, triangle NBM is similar to triangle ABC

But BC = 2BM....and since similar triangles are similar in all respects, AC = 2NM

But..triangle APC was shown to be similar to triangle MPN..... so.....if AC = 2MN, then PC = 2PN

And using the distance formula,  CN  = 6

So.....since PC + PN  = CN....then, by substitution,  2PN + PN = 6

...  so 3PN  = 6

....so PN = 2

And PC = 2PN = 2(2) =  4

Here's a pic :

CPhill  Jul 13, 2017
edited by CPhill  Jul 13, 2017
#3
+18281
+1

We have a right triangle triangle ABC where the legs AB and BC have lengths 6 and 3sqrt3, respectively.

Medians AM and CN meet at point P. What is the length of CP?

$$\begin{array}{rcll} \text{Let } \vec{a} &=& \vec{CB} \\ \text{Let } \vec{b} &=& \vec{AC} \\ \text{Let } \vec{c} &=& \vec{AB} = \vec{a} + \vec{b} \\ \text{Let } \vec{u} &=& \vec{CN} = \frac12\vec{c} - \vec{b} = \frac12(\vec{a} + \vec{b}) - \vec{b} = \mathbf{ \frac12 \vec{a} - \frac12 \vec{b} } \\ \text{Let } \vec{v} &=& \vec{AM} = \mathbf{ \vec{b} + \frac12 \vec{a} } \\ \end{array}$$

$$\begin{array}{|lrcll|} \hline \mathbf{ \vec{b}+\lambda \vec{u} - \mu \vec{v} } & \mathbf{=} & \mathbf{0} \\\\ & \vec{b}+\lambda \vec{u} - \mu \vec{v} &=& 0 \quad & | \quad \mathbf{ \vec{u} = \frac12 \vec{a} - \frac12 \vec{b} } \\ & \vec{b}+\lambda (\frac12 \vec{a} - \frac12 \vec{b}) - \mu \vec{v} &=& 0 \quad & | \quad \mathbf{ \vec{v} = \vec{b} + \frac12 \vec{a} } \\ & \vec{b}+\lambda (\frac12 \vec{a} - \frac12 \vec{b}) - \mu (\vec{b} + \frac12 \vec{a}) &=& 0 \\ & \vec{b}+ \frac{\lambda}{2} \vec{a} - \frac{\lambda}{2} \vec{b} - \mu \vec{b} - \frac{\mu}{2} \vec{a} &=& 0 \\ & \frac{\vec{a}}{2} \cdot (\underbrace{\lambda -\mu}_{=0}) &=& \vec{b} \cdot ( \underbrace{\frac{\lambda}{2} + \mu - 1}_{=0} ) \\\\ (1) & \lambda -\mu &=& 0 \\ & \mu &=& \lambda \\\\ (2) & \frac{\lambda}{2} + \mu - 1 &= & 0 \\ & \frac{\lambda}{2} + \lambda - 1 &= & 0 \\ & \frac{3}{2}\lambda &= & 1 \\ & \lambda &= & \frac{2}{3} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \overline{CP} &=& \lambda \cdot \overline{CN} \quad & | \quad \overline{CN}^2 = (3\cdot \sqrt{3})^2 + (\frac{6}{2})^2 = 36 \\ \overline{CP} &=& \frac{2}{3} \cdot 6 \quad & | \quad \overline{CN} = \sqrt{36} = 6 \\ \overline{CP} &=& 2\cdot 2 \\ \mathbf{\overline{CP}} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}$$

heureka  Jul 13, 2017
#4
+18281
0

We have a right triangle triangle ABC where the legs AB and BC have lengths 6 and 3sqrt3, respectively.

Medians AM and CN meet at point P.

What is the length of CP?

another approach

See labeling CPhill :

$$\begin{array}{rcll} \text{Let } \overline{AB} &=& 6 \\ \text{Let } \overline{BC} &=& 3 \sqrt{3} \\ \text{Let } \overline{CP} &=& ? \\\\ \text{Let } \vec{A} &=& \binom{0}{6} \\ \text{Let } \vec{B} &=& \binom{0}{0} \\ \text{Let } \vec{C} &=& \binom{ 3 \sqrt{3} }{0} \\ \end{array}$$

Point P is the barycenter of triangle ABC

$$\begin{array}{|rcll|} \hline \vec{P} &=& \frac13 \cdot( \vec{A} + \vec{B} + \vec{C} ) \\ \vec{P} &=& \frac13 \cdot\Big( \binom{0}{6} + \binom{0}{0} + \binom{ 3 \sqrt{3} }{0} \Big) \\ \vec{P} &=& \frac13 \cdot \binom{ 0+0+3\sqrt{3} } { 6+0+0 } \\ \vec{P} &=& \frac13 \cdot \binom{ 3\sqrt{3} } { 6 } \\ \vec{P} &=& \binom{ \sqrt{3} } { 2 } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \overline{CP} &=& |\vec{C}-\vec{P}| \\ &=& |\binom{ 3 \sqrt{3} }{0}-\binom{ \sqrt{3} } { 2 }| \\ &=& | \binom{ 3 \sqrt{3}-\sqrt{3} }{0-2} | \\ &=& | \binom{ 2\sqrt{3} }{-2} | \\ &=& \sqrt{ (2\sqrt{3})^2 + (-2)^2 } \\ &=& \sqrt{ 4\cdot 3 + 4 } \\ &=& \sqrt{ 16 } \\ \mathbf{\overline{CP} } & \mathbf{=} & \mathbf{4} \\ \hline \end{array}$$

heureka  Jul 13, 2017

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