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

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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"

Jul 4, 2015

#5
+18 $$\\\frac{sin \angle POM_1}{1}=\frac{sin45}{\sqrt5}\\\\ \frac{sin \angle POM_1}{1}=\frac{1}{\sqrt2*\sqrt5}\\\\ \frac{sin \angle POM_1}{1}=\frac{1}{\sqrt{10}}\\\\ \angle POM_1=asin\left( \frac{1}{\sqrt{10}}\right)\\\\$$

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

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

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

Now Find N

N is the intersection of

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

$$\underset{\,\,\,\,{{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\frac{{\mathtt{x}}}{{\sqrt{{\mathtt{3}}}}}}\\ {\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\\ {\mathtt{x}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{1.035\: \!276\: \!180\: \!410\: \!083}}\\ {\mathtt{x}} = {\mathtt{1.793\: \!150\: \!944\: \!336\: \!107}}\\ \end{array} \right\}$$

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

Distance ON

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

$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad correct to 3 decimal places$$$$${\mathtt{0.5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{11.565\: \!051\: \!177\: \!078}}^\circ\right)} = {\mathtt{0.464\: \!101\: \!615\: \!138\: \!509\: \!6}}$$ $$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad correct to 3 decimal places$$$

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

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   Jul 4, 2015
#2
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you got half of the anwsers,try it in another way,like using law of cosine

Jul 4, 2015
#3
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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!!!   Jul 4, 2015
#4
+5

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

∠OMP = 63.435o ,      ∠MOP = 71.565o ,    PM = 3

Jul 4, 2015
#5
+18 $$\\\frac{sin \angle POM_1}{1}=\frac{sin45}{\sqrt5}\\\\ \frac{sin \angle POM_1}{1}=\frac{1}{\sqrt2*\sqrt5}\\\\ \frac{sin \angle POM_1}{1}=\frac{1}{\sqrt{10}}\\\\ \angle POM_1=asin\left( \frac{1}{\sqrt{10}}\right)\\\\$$

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

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

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

Now Find N

N is the intersection of

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

$$\underset{\,\,\,\,{{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\frac{{\mathtt{x}}}{{\sqrt{{\mathtt{3}}}}}}\\ {\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\\ {\mathtt{x}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{1.035\: \!276\: \!180\: \!410\: \!083}}\\ {\mathtt{x}} = {\mathtt{1.793\: \!150\: \!944\: \!336\: \!107}}\\ \end{array} \right\}$$

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

Distance ON

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

$$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad correct to 3 decimal places$$$$${\mathtt{0.5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{2.070\: \!552\: \!360\: \!820\: \!166}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{11.565\: \!051\: \!177\: \!078}}^\circ\right)} = {\mathtt{0.464\: \!101\: \!615\: \!138\: \!509\: \!6}}$$ $$Area\;\triangle\;ONM_1= 0.464\;units^2 \qquad correct to 3 decimal places$$$

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

Melody Jul 4, 2015
#6
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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.  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. Jul 4, 2015
#7
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@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 30o , and the length of PM is 1, THEN the angle POM = 18.435o , and the area of a triangle NOM = 0.464u2 !

Bravo, Melody! Jul 4, 2015
#8
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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.....

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   Jul 4, 2015
#10
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the given in question one is not vaild, when you doing number two. Jul 4, 2015
#11
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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. Jul 4, 2015
#12
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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?

Jul 4, 2015
#13
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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. 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.

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 :)

Jul 5, 2015
#16
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Ya. Did you learn something from the anwser?

Jul 5, 2015
#17
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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. Jul 5, 2015
#18
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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. 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! Jul 5, 2015
#20
+5

Yes that is fine Fiora Jul 5, 2015