Angcheek Hypersonic October 3, 2015 Share October 3, 2015 Waaaa like that u also know... how long it took u ? ED = DC, Therefore △CDE is an isosceles triangle. Angle EDC = 90 + 60 =150 Angle CED = (180 - 150)/2 = 15 ? = 60 - 15 = 45 ↡ Advertisement 1 Link to post Share on other sites More sharing options...
1fast1 Supersonic October 8, 2015 Share October 8, 2015 calling the math pros @turboflat4, I need some help with the below problem - I know that the angles of the equilateral triangle are 60 each - and think that you should be able to work it out with either congruency or reciprocity - but I'm a little bit stuck - Any clues? GEOMETRY.jpg Sorry this is the first I'm seeing this, on a trip (actually been typing the last few posts from Frankfurt). Anyway, as @Ender has shown, this is a simple problem. But it immediately reminded me of a far more interesting one. Try this on for size: Problem : In the figure below, ABCD is a square. Prove that triangle DMC is equilateral. Remember this is a geometry problem. You can brute force it with trig, but then you're not being very clever. Link to post Share on other sites More sharing options...
Darryn Turbocharged October 8, 2015 Share October 8, 2015 Mab is isocles. Sorry this is the first I'm seeing this, on a trip (actually been typing the last few posts from Frankfurt). Anyway, as @Ender has shown, this is a simple problem. But it immediately reminded me of a far more interesting one. Try this on for size: Problem : In the figure below, ABCD is a square. Prove that triangle DMC is equilateral. equilateral_square_1.gif Remember this is a geometry problem. You can brute force it with trig, but then you're not being very clever. Link to post Share on other sites More sharing options...
Jamesc Hypersonic October 8, 2015 Share October 8, 2015 A business associate commented that Singaporeans are too inflexible. And I say this because our education system is a huge influence He wouldn't say that if he met these people. Link to post Share on other sites More sharing options...
Jamesc Hypersonic October 8, 2015 Share October 8, 2015 Crap, I knew I was missing something about one of the properties of the shape - Didn't think it would be that easy though! Thanks Why didn't you just measure it with a protractor? How hard can it? Link to post Share on other sites More sharing options...
Jamesc Hypersonic October 8, 2015 Share October 8, 2015 I really worry about the parents these days. Everything also must ask someone else and everything also don't know how to do themselves. Link to post Share on other sites More sharing options...
Angcheek Hypersonic October 8, 2015 Share October 8, 2015 Thats bec the subjects are getting more difficult and what they learnt from sckool already give back to teachers . I really worry about the parents these days. Everything also must ask someone else and everything also don't know how to do themselves. Holiday hur ? Or go drink free beer ? Sorry this is the first I'm seeing this, on a trip (actually been typing the last few posts from Frankfurt). Anyway, as @Ender has shown, this is a simple problem. But it immediately reminded me of a far more interesting one. Try this on for size: Problem : In the figure below, ABCD is a square. Prove that triangle DMC is equilateral. equilateral_square_1.gif Remember this is a geometry problem. You can brute force it with trig, but then you're not being very clever. 2 Link to post Share on other sites More sharing options...
Jamesc Hypersonic October 8, 2015 Share October 8, 2015 Brute force it with trig is for losers. Just measure the 3 sides with a ruler lah. If all 3 side is equal then its an equal lateral triangle lah. So easy. QED Quite Easily Done. 3 Link to post Share on other sites More sharing options...
ST69 4th Gear October 8, 2015 Share October 8, 2015 Sorry this is the first I'm seeing this, on a trip (actually been typing the last few posts from Frankfurt). Anyway, as @Ender has shown, this is a simple problem. But it immediately reminded me of a far more interesting one. Try this on for size: Problem : In the figure below, ABCD is a square. Prove that triangle DMC is equilateral. equilateral_square_1.gif Remember this is a geometry problem. You can brute force it with trig, but then you're not being very clever. looks like ADM & BCM also isosceles triangle, hence DMC is equilateral. Link to post Share on other sites More sharing options...
1fast1 Supersonic October 8, 2015 Share October 8, 2015 looks like ADM & BCM also isosceles triangle, hence DMC is equilateral. Since when did any math teacher accept "looks like"? Got to prove it bro. Mab is isocles. Yes with base angles of 15 degrees. That's obvious, cos it's given. How do you show DMC is equilateral? Thats bec the subjects are getting more difficult and what they learnt from sckool already give back to teachers . Holiday hur ? Or go drink free beer ? No lah. Anyway Oktoberfest actually mostly happens in Sep and is generally over by the third of October. Never went to Munich anyway. I went to the Ring dude. More on that later. Difficult to type from phone and tired. Anyway flight out tomorrow. Brute force it with trig is for losers. Just measure the 3 sides with a ruler lah. If all 3 side is equal then its an equal lateral triangle lah. So easy. QED Quite Easily Done. Haha. https://en.m.wikipedia.org/wiki/Missing_square_puzzle Careful with "Quite Easily Done", let's stick to "Quod Erat Demonstrandum". 1 Link to post Share on other sites More sharing options...
Porker Turbocharged October 8, 2015 Share October 8, 2015 Since when did any math teacher accept "looks like"? Got to prove it bro. Yes with base angles of 15 degrees. That's obvious, cos it's given. How do you show DMC is equilateral? No lah. Anyway Oktoberfest actually mostly happens in Sep and is generally over by the third of October. Never went to Munich anyway. I went to the Ring dude. More on that later. Difficult to type from phone and tired. Anyway flight out tomorrow. Haha. https://en.m.wikipedia.org/wiki/Missing_square_puzzle Careful with "Quite Easily Done", let's stick to "Quod Erat Demonstrandum". Deh pundek please stop spamming this forum with your Caltech'ish' mathematics theorems and such can 😂 Link to post Share on other sites More sharing options...
1fast1 Supersonic October 8, 2015 Share October 8, 2015 Deh pundek please stop spamming this forum with your Caltech'ish' mathematics theorems and such can 😂 Better than you spamming your gay cum everywhere. ASS-ume I attached an image of Hard Gay here. Too tired to find and do it on a phone. Link to post Share on other sites More sharing options...
Porker Turbocharged October 8, 2015 Share October 8, 2015 Better than you spamming your gay cum everywhere. ASS-ume I attached an image of Hard Gay here. Too tired to find and do it on a phone. Wait till you're back then I spam on you lah. Miss you deep pundek 😂 Link to post Share on other sites More sharing options...
ST69 4th Gear October 9, 2015 Share October 9, 2015 (edited) True true...was lazy to type it out lah...But for you..., The side isoceles triangles angle (90-15) =75. Bottom isosceles angle (180-15-15) = 150. The centre triangle angle is 360-150-75-75 = 60, hence equilateral triangle lah... (thot it was self explanatory... Since when did any math teacher accept "looks like"? Got to prove it bro. Edited October 9, 2015 by ST69 Link to post Share on other sites More sharing options...
1fast1 Supersonic October 9, 2015 Share October 9, 2015 True true...was lazy to type it out lah...But for you..., The side isoceles triangles angle (90-15) =75. Bottom isosceles angle (180-15-15) = 150. The centre triangle angle is 360-150-75-75 = 60, hence equilateral triangle lah... (thot it was self explanatory... How do you know (prove) the side triangle is isosceles? You're making a large assumption there. Only the base of triangle DMC can be assumed to be the side length of the square. The other two sides are not given. If you assume that "side" triangle is isosceles then you don't even need all that. The other two lengths of DMC are now also equal to its base so it's equilateral. No need to even consider angles. But the point is: you have to show the side triangle is isosceles in the first place. Link to post Share on other sites More sharing options...
Muddy 1st Gear October 9, 2015 Share October 9, 2015 Sorry this is the first I'm seeing this, on a trip (actually been typing the last few posts from Frankfurt). Anyway, as @Ender has shown, this is a simple problem. But it immediately reminded me of a far more interesting one. Try this on for size: Problem : In the figure below, ABCD is a square. Prove that triangle DMC is equilateral. equilateral_square_1.gif Remember this is a geometry problem. You can brute force it with trig, but then you're not being very clever. One possible way is to assume that it is an equilateral triangle and verify that none of the properties are broken. If it is true, that it is proven. Angle ADM + CDM = Angle BCM + DCM = 90 and Circle of origin M = Angle AMB + BMC + CMD + AM = 360 Link to post Share on other sites More sharing options...
1fast1 Supersonic October 9, 2015 Share October 9, 2015 One possible way is to assume that it is an equilateral triangle and verify that none of the properties are broken. If it is true, that it is proven. Angle ADM + CDM = Angle BCM + DCM = 90 and Circle of origin M = Angle AMB + BMC + CMD + AM = 360 Afraid that's mathematically unsound. You have to go in the direction of the implication, i.e. From the 15 degrees to the equilateral triangle. Have to cut this post short cos leaving for airport but suffice it to say this is not a trivial problem unlike what was posted before. I've already sketched out a solution and will post the images on my return if no one has solved it in that time. Link to post Share on other sites More sharing options...
Darryn Turbocharged October 9, 2015 Share October 9, 2015 Afraid that's mathematically unsound. You have to go in the direction of the implication, i.e. From the 15 degrees to the equilateral triangle. Have to cut this post short cos leaving for airport but suffice it to say this is not a trivial problem unlike what was posted before. I've already sketched out a solution and will post the images on my return if no one has solved it in that time. I'm thinking of playing around with vertical and horizontal bisectors - wondering if that would help at all I also notice that the angles are "fun" (i.e 15 / 75 / 150 / 60) - all are multiples of 15 - and wondering if this has some sort of impact ↡ Advertisement Link to post Share on other sites More sharing options...
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