A Nice geometry math question for competitive exams| math olympiad| Harvard University exam

This is not drawn to scale and this is not an easy one. Yes, welcome back to my channel. We have a lovely geometry question to solve. Please this is not drawn to scale. In this question we have a right angle triangle and inside this right angle a circle is inscribed and we are asked to find a P which happens to be the diameter of this circle. Okay first of all let's find the center of this circle. This is the center. Let's assume this is the center because this is a freehand sketch. So if this is the center of this circle, remember that any line, any straight line from the center of the circle to its circumference is the radius of that circle. Okay and we have important points here. These two points where this line and the circle touches. That means EF is a tangent. So when we draw a straight line from there to the center of this circle we are going to have another radius but this particular radius is special because we are going to form right angle. Okay I'm going to explain later. So similarly if we take it to this point again we form another right angle. All right so in a circle of center zero and a tangent AB. Okay this is a tangent please. If you drop a straight line from the center to the point of contact between the circle and the tangent, 90 degrees is formed. So that is why at this point we have 90 degrees here and we have another 90 degrees there. Okay now note that this is O right. OB, OD is also the radius of this circle. OA is the radius as well. OB is the radius as well and AB is the diameter which is what we are looking for. Now if you look closely you will notice that triangle EOC is similar to triangle ODG. Okay now if two triangles are similar let's prove that they are similar. For them to be similar the angles has to be equal right. Okay now we have 90 degrees here and we have 90 degrees here. So angle C is equal to angle D. Check. All right now we can also see that angle E is equal to angle O. Check. Now let's explain that. Remember that FE is parallel to OD. So we have something like this whereby this is FE and this is OD. Okay this is the O here just extrapolated. Now they are parallel to each other and we have a line from here like this. That means EG is a diagonal, sorry is a transversal. Okay so EG is transversal. Now if it is a transversal then remember that corresponding angles are equal. So it simply means that this angle here will be equal to this angle here. They are corresponding angles. So that proves that angle E is equal to angle O. Now if we have two angles equal it simply means that the other two are equal as well. So that proves that triangle EOC is similar to triangle ODG. Now when two triangles are similar the ratio of their corresponding sides are equal. So let's call this X and call this Y. Okay it means that X over R is equal to R over Y. If we cross multiply cross multiply we have R times R is R squared and X times Y is XY. Now let's call this equation one. Now if we look at triangle EOC it is a right angle triangle. So EO squared should be equal to EC squared plus CO squared. All right so EO is what 3 plus R. It should be equal to EC that is X plus CO that is R. Now remember that A plus B squared is equal to A squared plus 2AB plus B squared. So it means that this is 3 squared plus 2 multiplied by 3 multiplied by R plus R squared is equal to this X squared plus R squared. So 3 squared is 9 plus 2 times 3 times R is 6R plus R squared is equal to X squared plus R squared. Now plus R and plus R squared since they are on different sides of the equation they will definitely cancel out because if we transfer this to this side becomes minus R squared and plus R squared minus R squared is zero. So we are left with 9 plus 6R is equal to X squared. Now let's call this equation 2. Excuse me I need to be sure this is still on. Okay I'm back it's still on. Now let's continue. Similarly okay so permitting to write this here that X squared is equal to 9 plus 6R. This is equation 2. All right now let's find Y squared. So similarly in triangle ODG. This is also a right angle. OG squared should be equal to OD squared plus DG squared. So OG is R plus 8. R squared is equal to OD which is R plus DG which is Y. So if we expand this we will have R squared plus 2 times 8 times R is 16R plus 8 squared which is 64 is equal to R squared plus Y squared. R squared you remove R squared and we have that Y squared is equal to 16R plus 64 and this is equation 3. Now these equations are very very important. Let me show you how we will use it to find the value of our diameter. So if we substitute equation 2 and equation 3 into equation 1. Remember that equation 1 says R squared equals XY. Now let's square both sides. We will have R squared squared is equal to XY squared. So remember that A to the power of M raised to the power of N is equal to A to the power of M times N. Okay so let me write this somewhere because I need to Alright so that simply means that this is R to the power of 4 and this is X squared multiplied by Y squared. So we are going to substitute equation 2 and equation 3 into this. So we will have R to the power of 4 is equal to X squared is 9 plus 6R and Y squared is 16R plus 64. Now let's open the bracket. So R to the power of 4 is equal to. Let me make use of my calculator because I want this to be faster. I don't want the video to be too long. Okay okay so we have 9 times 16. That is 144. So we have 144R plus 9 times 64. 576 plus 6 times 16. 96. So it will be 96R squared and 6R times 64. We have 384 plus 384R. Okay alright now we collect like terms. We have R to the power of 4 is equal to 144 plus 384. That is 528. So we have 528R plus 96R squared plus 576. So you can see that we have a polynomial here of degree 4. Right so let us solve it. We are going to not solve all of them. We just need a value from there and we are done. Yes have you given us a thumbs up yet? If not please do that. Okay my video cut here and I didn't know I was just shooting the video until the end only for me to find out that at this point the video tripped off. So we are going to continue from here. Now we can see that let's transfer everything to this side. We will now have R to the power of 4 is equal to sorry we are going to bring in 9R squared since we don't have R to the power of 3. So this will come to become minus 96R squared. Okay and this becomes minus 528R. This becomes minus 576. Everything is equal to zero. So I'm going to copy this again. So what we have is R to the power of 4 minus 96R squared minus 528R minus 576 is equal to zero. So here you can see that we have a polynomial and this polynomial is of the fourth degree. So to solve it let's use the synthetic division method. Now in synthetic division method we are going to first of all use trial and error method to find the first root of this equation. Okay all right then we now use it to synthetically divide the particular equation to get the ahead. So first of all by trial and error the first root is 12. Okay so it means that R is equal to 12 right which implies that R minus 12 is what a root is a factor. Now using the synthetic division method we'll write the coefficient of everything here. So coefficient of R to the power of 4 is 1. We're supposed to have R to the power of 3 but it's not here. It means that its coefficient is 0. So we have 0 we have minus 96 we have minus 528 and we have minus 576. So we are dividing by 12. We'll bring this down to 1. 12 times 1 is 12. 0 plus 12 is 12. 12 times 12 is 144. Now 144 minus 96 we need to use my calculator so that we don't waste much time. Okay yeah so using calculator we have 144 minus 96 which is 48. So we have sorry the 48 should be here. So 12 times 48 is 576. So we subtract 576 minus 528. We have 48 again and 12 times 48 gives us 576. And when we subtract this minus this plus this we give us 0. So this since the remainder is 0 it means that R minus 12 is actually a factor. All right now we are going to write we will write R minus 12 bracket then this will be R to the power of 3 okay then plus 12 R squared plus 48 R plus 48. Now everything is equal to 0. So this second equation is going to this second factor is going to be in cubic okay since this is in fourth degree. So from here we can see that R minus 12 is 0 and R is equal to 12. Now we don't need to stress ourselves further solving this. We've gotten our R to be 12 which is a very reasonable answer. All right now what we now do is it simply means that R which happens to be the radius of this circle is 12. And we are asked to find AB and AB is R plus R. So it means that AB is 12 plus 12. So our AB is 24. Thank you for watching and see you in my next video.