In the following image, triangle OPQ is an isosceles right triangle;the measure of angle POQ is 90 degrees,

OP= $${\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}$$, point M is on segment PQ.

(1)If $${\mathtt{OM}} = {\sqrt{{\mathtt{5}}}}$$,what is the length of PM?

(2)If point N is on segment OP,and the measure of angle NOP is 30 degrees

Question:What is the measure of angle POM while triangle NOM has its possible mininum area ?

and find out mininum area of triangle NOM.

remenber "IF"

fiora Jul 4, 2015

#5**+18 **

$$\\\frac{sin \angle POM_1}{1}=\frac{sin45}{\sqrt5}\\\\

\frac{sin \angle POM_1}{1}=\frac{1}{\sqrt2*\sqrt5}\\\\

\frac{sin \angle POM_1}{1}=\frac{1}{\sqrt{10}}\\\\

\angle POM_1=asin\left( \frac{1}{\sqrt{10}}\right)\\\\$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\sqrt{{\mathtt{10}}}}}}\right)} = {\mathtt{18.434\: \!948\: \!822\: \!922^{\circ}}}$$

$$\\\angle NOM_1=30-18.434948822922=\;11.565051177078$$

$${\mathtt{30}}{\mathtt{\,-\,}}{\mathtt{18.434\: \!948\: \!822\: \!922}} = {\mathtt{11.565\: \!051\: \!177\: \!078}}$$

Now Find N

N is the intersection of

$$y=(tan30)x = \frac{x}{\sqrt3} \qquad and \qquad y=-x+2\sqrt2$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\frac{{\mathtt{x}}}{{\sqrt{{\mathtt{3}}}}}}\\

{\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\\

{\mathtt{x}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\\

\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{1.035\: \!276\: \!180\: \!410\: \!083}}\\

{\mathtt{x}} = {\mathtt{1.793\: \!150\: \!944\: \!336\: \!107}}\\

\end{array} \right\}$$

$$N(3\sqrt2-\sqrt6,\;\sqrt6-\sqrt2)$$

Distance ON

$${\sqrt{{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{6}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{6}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}$$

$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad $correct to 3 decimal places$$$

$${\mathtt{0.5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{11.565\: \!051\: \!177\: \!078}}^\circ\right)} = {\mathtt{0.464\: \!101\: \!615\: \!138\: \!509\: \!6}}$$

$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad $correct to 3 decimal places$$$

NOW If I haven't made any stupid errors (which I probably have) that should be correct :)

Melody Jul 4, 2015

#1**+13 **

Here's (1)

sin MPO/OM = sin PMO/ PO

sin(45)/√5 = sin PMO /[2√2]

1/[√2 * √5] = sin PMO /[2√2]

1/√5 = sin PMO / 2.......therefore

sin^{-1}( 2/√5) = PMO = about 63.43°

Then angle MOP = [180 - 45 - 63.43]° = about 71.57º

So

√5/sin 45 = PM/sin71.57

PM = √5sin71.57 / sin45 = 3

CPhill Jul 4, 2015

#3**+10 **

OK, Fiora........

OM^2 = PM^2 + OP^2 - 2(PM)(OP)sin 45

5 = PM^2 + 8 - 2(PM)(2√2)(1/√2)

-3 = PM^2 - 4PM

PM^2 - 4PM + 3 = 0

(PM - 1)(PM - 3) = 0

PM = 1, 3

Is that correct ???

I see where I didn't take into account the SSA situation in my first answer!!!

CPhill Jul 4, 2015

#4**+5 **

The angle NOP is NOT 30^{o} ; but if it is, then a point N is in between M and P.

∠OMP = 63.435^{o} , ∠MOP = 71.565^{o} , PM = 3

civonamzuk Jul 4, 2015

#5**+18 **

Best Answer

$$\\\frac{sin \angle POM_1}{1}=\frac{sin45}{\sqrt5}\\\\

\frac{sin \angle POM_1}{1}=\frac{1}{\sqrt2*\sqrt5}\\\\

\frac{sin \angle POM_1}{1}=\frac{1}{\sqrt{10}}\\\\

\angle POM_1=asin\left( \frac{1}{\sqrt{10}}\right)\\\\$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\sqrt{{\mathtt{10}}}}}}\right)} = {\mathtt{18.434\: \!948\: \!822\: \!922^{\circ}}}$$

$$\\\angle NOM_1=30-18.434948822922=\;11.565051177078$$

$${\mathtt{30}}{\mathtt{\,-\,}}{\mathtt{18.434\: \!948\: \!822\: \!922}} = {\mathtt{11.565\: \!051\: \!177\: \!078}}$$

Now Find N

N is the intersection of

$$y=(tan30)x = \frac{x}{\sqrt3} \qquad and \qquad y=-x+2\sqrt2$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\frac{{\mathtt{x}}}{{\sqrt{{\mathtt{3}}}}}}\\

{\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\\

{\mathtt{x}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\\

\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{1.035\: \!276\: \!180\: \!410\: \!083}}\\

{\mathtt{x}} = {\mathtt{1.793\: \!150\: \!944\: \!336\: \!107}}\\

\end{array} \right\}$$

$$N(3\sqrt2-\sqrt6,\;\sqrt6-\sqrt2)$$

Distance ON

$${\sqrt{{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{6}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{6}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}$$

$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad $correct to 3 decimal places$$$

$${\mathtt{0.5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{11.565\: \!051\: \!177\: \!078}}^\circ\right)} = {\mathtt{0.464\: \!101\: \!615\: \!138\: \!509\: \!6}}$$

$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad $correct to 3 decimal places$$$

NOW If I haven't made any stupid errors (which I probably have) that should be correct :)

Melody Jul 4, 2015

#6**0 **

Hey,Melody!

I had mention that remenber "If" at the end of my question.That mean the given in question one is not vaild, when you doing number two.

I guess a few picture can help you understand my question.

Angle NOM is moving around in triangle QOP,what is the possible mininum area of triangle NOM?and find the measure of angle MOP.Sorry Melody.

fiora Jul 4, 2015

#7**+10 **

@fiora:/ Melody's answered both questions of yours.

1st answer: The length of PM could be either 1 or 3 ;

2nd answer: IF a point N is on a line segment PQ, the angle NOP is 30^{o} , and the length of PM is 1, THEN the angle POM = 18.435^{o} , and the area of a triangle NOM = 0.464u^{2} !

Bravo, Melody!

civonamzuk Jul 4, 2015

#8**+5 **

No.Here is a graph from desmos:

https://www.desmos.com/calculator/a5dnftk4qi

move a

I checked Melody's anwser.She is right if my question like that,however it's not.....

fiora Jul 4, 2015

#9**+5 **

I do not understand what you are trying to tell me Fiora.

(Neither does Civonamzuk or CPhill)

Perhaps there is a translation problem?

M can only be those 2 set points.

And you have said that angle PON=30 degrees.

So if N lies on the interval PQ, there is only one point that N it can be.

Thanks for my points

Melody Jul 4, 2015

#10**+5 **

the given in question one is not vaild, when you doing number two.

see my link on demos,https://www.desmos.com/calculator/a5dnftk4qi

fiora Jul 4, 2015

#11**+5 **

Like I said, I think there must be a translation problem :)

It doesn't matter, if you understand that is good.

I did look at your graph but so far it does not make sense to me.

Melody Jul 4, 2015

#12**0 **

Okay,Melody.At first I would like to apologize for that.And thank you for spend your time to solve the question.

Now,I have to explain my question.I stated "if" at the begin of each questions.That mean you can not use the given in qustion 1 to solve qustion 2.

But if I did state "if" at the begin of the qustion ,like "$${\mathtt{OM}} = {\sqrt{{\mathtt{5}}}}$$,what is the length of PM?",then you can use it to solve question 2. and I stated if,so.......

Now,if you could only use the given in number two and the title,can you do qustion 2?

fiora Jul 4, 2015

#13**0 **

Hi Fiora,

I was not in the slightest little bit offended. So you certainly do not need to appologize.

I was just a little confused. I still am.

Well if M is allowed to be any point on QP then why not make M approach N.

In this case the area of the triangle NOM would approach 0. You can't get much smaller than that.

Melody Jul 5, 2015

#14**+10 **

No,the question given that the measure of angle NOP is 30 degrees.

Indeed ,this is a translation problem.

Acctually,this question was created by chinese.I translate it and post it on the forum.Yes,I am a chinese.

See,

I guess you dont want to do this question anymore ,so I post the anwsers

Math is an international language.

fiora Jul 5, 2015

#15**+5 **

Thank you Fiora,

Yes maths is an international language but the words that are used around the maths are not :)

Melody Jul 5, 2015

#17**0 **

I had done this before I post it in the forum.I post this because I think you guys could learn something from this question.If you don't,I will not post this kind question anymore.

fiora Jul 5, 2015

#18**0 **

Hello Fiora,

None of us mind you posting questions like this, it is good that the forum is used in a variety of ways and your challeges are good ones.

But I think it is polite for you to state that you know the answer and that you are posting it as a challenge to other people if they are interested.

I was very tired last night and spent a long time on your question because I thought you were waiting for an answer. If I had known that I was doing it only for my own education I would have chosen to use my time differently.

It doesn't matter though, next time I will know.

Melody Jul 5, 2015

#19**+5 **

Thank you,Melody!

I usually try to figure out anwsers of my questions before I post it into the forum.I am having fun with doing challenges question like this.

I will not present like "I had done this" in my question,but I will indicatet that if I dont know how to do my question.So,you could do my question when you have time.Is that okay? Thank you!

fiora Jul 5, 2015