Stage+1,+2+of+our+PM2+project--instuctions+and+answers

These are the things that we have to work on in stage 1 and 2: =﻿= If any of U guys wants to edit something, copy and paste the answer that you want to edit, then edit it. please remember to indicate your name behind every edit. Please be honest. ~ Jia Wei

=__**Stage 1:**__= ==Prove that the right isoceles triangle in the circle's area is larger than any other isoceles triangle's area. The base must be the same as the diameter of the circle. (The last sentence is what I thought I heard... Jia Wei)== =﻿= =Answer: When the base is the same (diameter of the circle), the isosceles triangles have a larger height than the scalene triangles. ~ Jia Wei= =Yup. That is definitely so. I also checked it and found it to be true. But i don't know if that is the proof...= =﻿~ Tyrone= =﻿= =﻿= =﻿=

=﻿= =__**St﻿age 2:**__= ==﻿﻿﻿Prove that when an isoceles triangle has **ONE** same angle with another triangle in the circle, (the angle cannot any of the angles that are similar to one another because the remaining two angles might be the same) the isoceles triangle is always the larger triangle (the isoceles triangle cannot share a same angle with the equilateral triangle as an isoceles triangle cannot have an angle with 60 degrees).== =Possible Answer (Not confirmed) : An isosceles triangle is half of a square while a scalene triangle is half a rectangle, which is smaller if the sum of two sides forming the right angle adds up to the same number. Since the triangles are in the same circle, the sides forming the right angle must add up the same numbers. (I don't know whether this is true or not) ~ Jia Wei= =﻿= =An isosceles triangle may also be half of a rectangle, you know? (And your investigation conclusion does seem weird) I'll try to check it whenever I have time in the weekend. ~ Tyrone= =I don't think so. Since the isosceles triangle has two sides that have the same length, it must be half of a square. (I think so...) Example: If a square has a same perimeter as a rectangle, the square's area is always the biggest. If you don't believe it, try it... The rule applies if the area of the square and the rectangle is halved, in this case, it is a triangle. ~ Jia Wei= =﻿= =Isn't a thin (measuring across) isosceles triangle half of a rectangle? (pls check ur email, Jia Wei, i sent the mentioned triangle to you by email) ~ Tyrone= =﻿= =Tyrone, if the angle that is the same is a 90 degrees, then the isosceles is half of a square, if the angle that is the same is not a 90 degrees, then is is half a rectangle, but I think that if the isosceles triangle is half of the rectangle, then the rectangle's area is bigger than the rectangle which is the scalene triangle times 2 (know what I mean?). Then I have a weird thought: The equilateral triangle have the biggest perimeter (is that so?). By the way, I think that and equilateral is definitely half of a rectangle, because, suppose the sides of the equilateral triangle is 5cm. The two sides that are slanted cannot have a height that is higher then 5cm (because if two of the length are the same and one of them is slanted, it cannot be longer then the one that is straight. Also it is the main reason why a rhombus's that have the same side as a square can never have an area that is bigger than the square). It is purely because the sum of the sides of the rectangle is much bigger than that of a square that the area is bigger than the largest possible right-angle triangle that can be drawn in a circle (reason is stated above). ~ Jia Wei= =﻿= =﻿Jia Wei, I agree with your investigation, and I fully understand what you mean. Did you do anymore investigation? I still think that the isosceles triangle in the circle's area should be larger than the scalene triangle in the circle's area. And I agree that the rhombus is smaller than a square, but what has the rhombus got to do with our project? ~ Tyrone= =﻿= =I just took it as an example to proof that a straight line is longer that the slanted one, it may help in the later stages. I thought we finished this stage already? ~ Jia Wei= =﻿= =﻿We can still work on it, and anyway I thought we haven't proved it? I thought you only gave the example? (and okay, I understand your bringing the concept into the picture) ~ Tyrone= =﻿= =Oh yeah... We haven't proved that ' Since the triangles are in the same circle, the sides forming the right angle must add up the same numbers. ' is correct... ~ Jia Wei= =**I was thinking that triangles are halves of a squares and rectangles. So, I thought maybe cutting and pasting the triangles into the quadrilaterals might help. A triangle that has the circle as the base will have the area that is equal to (in this case, r** radius) r x r / 2. An equilateral triangle will have a ratio of height / base that is 866 / 1000 (use sine). The proof that the triangles in the same circle will have the same sum when the angles forming the right angle is added up is that, when you shorten one side of the two sides that from the 90 degrees, you can shift the shortened line done, thus giving space for the other line to extend (and vice versa). ~ Jia Wei= =﻿= =Okay, ﻿I understand, did you work that out by yourself? (and I checked it and found it to be correct, except the 866 / 1000 part I didn't check) ~ Tyrone=

=Of course, if not I would not have posted it or I will at least indicate the person behind the part that I found out.=

=Actual Question :=

=This is what I have done:=

=I drew out the degrees : 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330, 360. Then I connected the points 60, 120, and 360. I found out that the height of the equilateral triangle is 3 quarters of the diameter of the circle. Dividing the triangle into 9 equal parts ( 5 at the bottom, 3 in the middle, and 1 at the top ), I arrived at the conclusion above. Using trigonometry ( reverse this time you need to use the height of the triangle to calculate the base of the triangle ). Using these I think you guys can finish the project right? Based on the information above the supposed formula is (square-root ﻿the area of the circle over pie time 2 over 3) times (square-root the area of the circle of the pie times 2 over 3 again times 1000 over 866) over 2. Maybe trigonometry can proof that equilateral triangle is the biggest by calculating the ratio of the height of the triangle and the base of the triangle. Also, try to find the relationship of the height or the base of the triangles to the diameter or radius of the circle. The formula (which I know is not correct, but the answer will be very near) can be simplified as : (Radius of the circle times 2/3) times (Radius of the circle times 2/3 times 866/1000) over 2. If you draw the 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330, 360 degrees of a straight line in a circle you will find that the height of any equilateral triangle in a circle will be 2/3 of the diameter. So, based on that and trigonometry, you can find the formula to find the area of the largest possible equilateral triangle that can be drawn in the circle. You guy's job is to proof that the height of the triangle is 2/3 the diameter of the circle and the equilateral triangle __**//is//**__ the largest triangle inside a circle. ~ Jia Wei = =The triangles are as follows:= =One of the angles can be:= = ﻿ ﻿Degree(s): =
 * =Degrees= || =Ratio of the triangle's height over the base= ||
 * =1= || =0.017= ||
 * =2= || =0.034= ||
 * =3= || =0.053= ||
 * =4= || =0.070= ||
 * =5= || =0.087= ||
 * =6= || =0.105= ||
 * =7= || =0.122= ||
 * =8= || =0.140= ||
 * =9= || =0.157= ||
 * =10= || =0.175= ||
 * =11= || =0.193= ||
 * =12= || =0.210= ||
 * =13= || =0.228= ||
 * =14= || =0.246= ||
 * =15= || =0.263= ||
 * =16= || =0.281= ||
 * =17= || =0.299= ||
 * =18= || =0.317= ||
 * =19= || =0.335= ||
 * =20= || =0.353= ||
 * =21= || =0.371= ||
 * =22= || =0.389= ||
 * =23= || =0.407= ||
 * =24= || =0.425= ||
 * =25= || =0.222= ||
 * =26= || =0.462= ||
 * =27= || =0.480= ||
 * =28= || =0.499= ||
 * =29= || =0.517= ||
 * =30= || =0.536= ||
 * =31= || =0.555= ||
 * =32= || =0.562= ||
 * =33= || =0.592= ||
 * =34= || =0.611= ||
 * =35= || =0.631= ||
 * =36= || =0.650= ||
 * =37= || =0.669= ||
 * =38= || =0.689= ||
 * =39= || =0.708= ||
 * =40= || =0.728= ||
 * =41= || =0.748= ||
 * =42= || =0.768= ||
 * =43= || =0.788= ||
 * =44= || =0.808= ||
 * =45= || =0.828= ||
 * =46= || =0.849= ||
 * =47= || =0.870= ||
 * =48= || =0.890= ||
 * =49= || =0.911= ||
 * =50= || =0.933= ||
 * =51= || =0.954= ||
 * =52= || =0.975= ||
 * =53= || =0.997= ||
 * =54= || =1.019= ||
 * =55= || =1.041= ||
 * =56= || =1.063= ||
 * =57= || =1.086= ||
 * =58= || =1.109= ||
 * =59= || =1.132= ||
 * =60= || =1.155= ||
 * =61= || =1.178= ||
 * =62= || =1.202= ||
 * =63= || =1.226= ||
 * =64= || =1.250= ||
 * =65= || =1.274= ||
 * =66= || =1.299= ||
 * =67= || =1.324= ||
 * =68= || =1.349= ||
 * =69= || =1.375= ||
 * =70= || =1.400= ||
 * =71= || =1.178= ||
 * =72= || =1.453= ||
 * =73= || =2.108= ||
 * =74= || =1.507= ||
 * =75= || =1.535= ||
 * =76= || =1.563= ||
 * =77= || =1.591= ||
 * =78= || =1.620= ||
 * =79= || =1.649= ||
 * =80= || =1.678= ||
 * =81= || =1.708= ||
 * =82= || =1.739= ||
 * =83= || =1.769= ||
 * =84= || =1.801= ||
 * =85= || =1.833= ||
 * =86= || =1.865= ||
 * =87= || =1.898= ||
 * =88= || =1.931= ||
 * =89= || =1.965= ||
 * =90= || =2= ||
 * =91= || =2.035= ||
 * =92= || =2.071= ||
 * =93= || =2.108= ||
 * =94= || =2.145= ||
 * =95= || =2.183= ||
 * =96= || =2.221= ||
 * =97= || =2.261= ||
 * =98= || =2.301= ||
 * =99= || =2.342= ||
 * =100= || =2.384= ||
 * =101= || =2.426= ||
 * =102= || =2.470= ||
 * =103= || =2.514= ||
 * =104= || =2.560= ||
 * =105= || =2.606= ||
 * =106= || =1.327= ||
 * =107= || =2.703= ||
 * =108= || =2.753= ||
 * =109= || =2.804= ||
 * =110= || =2.856= ||
 * =111= || =2.910= ||
 * =112= || =2.965= ||
 * =113= || =3.022= ||
 * =114= || =3.080= ||
 * =115= || =3.139= ||
 * =116= || =3.201= ||
 * =117= || =3.264= ||
 * =118= || =3.329= ||
 * =119= || =3.395= ||
 * =120= || =3.464= ||
 * =121= || =3.535= ||
 * =122= || =3.608= ||
 * =123= || =3.684= ||
 * =124= || =3.761= ||
 * =125= || =3.842= ||
 * =126= || =3.925= ||
 * =127= || =4.011= ||
 * =128= || =4.101= ||
 * =129= || =4.193= ||
 * =130= || =4.289= ||
 * =131= || =4.389= ||
 * =132= || =4.492= ||
 * =133= || =4.600= ||
 * =134= || =4.712= ||
 * =135= || =4.828= ||
 * =136= || =4.950= ||
 * =137= || =5.077= ||
 * =138= || =5.210= ||
 * =139= || =5.349= ||
 * =140= || =5.495= ||
 * =141= || =5.648= ||
 * =142= || =5.808= ||
 * =143= || =2.989= ||
 * =144= || =6.155= ||
 * =145= || =6.343= ||
 * =146= || =6.542= ||
 * =147= || =6.752= ||
 * =148= || =6.975= ||
 * =149= || =7.212= ||
 * =150= || =7.464= ||
 * =151= || =7.733= ||
 * =152= || =8.022= ||
 * =153= || =8.331= ||
 * =154= || =8.663= ||
 * =155= || =9.021= ||
 * =156= || =9.401= ||
 * =157= || =9.830= ||
 * =158= || =10.289= ||
 * =159= || =10.698= ||
 * =160= || =11.343= ||
 * =161= || =11.952= ||
 * =162= || =12.628= ||
 * =163= || =13.382= ||
 * =164= || =14.231= ||
 * =165= || =15.192= ||
 * =166= || =16.289= ||
 * =167= || =17.554= ||
 * =168= || =19.029= ||
 * =169= || =20.771= ||
 * =170= || =22.860= ||
 * =171= || =25.412= ||
 * =172= || =28.061= ||
 * =173= || =32.700= ||
 * =174= || =38.162= ||
 * =175= || =45.808= ||
 * =176= || =57.273= ||
 * =177= || =76.377= ||
 * =178= || =114.580= ||
 * =179= || =229.177= ||



=//This is the GRAND FINALI! //=

=//**The relationship between the diameter of the circle and the largest triangle is : 3/4 of the diameter is the height of the triangle. Proof: Draw the radius of the circle, (it can be any line as long as it is straight). Then, draw a line branching out of the original line at exactly 120 degrees and has the same length. Then, extend the original line, then draw a horizontal line connecting the extended part of the line and the second line, forming a triangle. This triangle's angles are, 30, 60, 90. Using cos, we can find that the original line is twice the length of the extended line. Since the second line has the same length as the first line, which is the radius of the circle, it can be concluded the the extended part, which is the height of the triangle, is 3/4 of the diameter of the circle.**//=