Thanks for the video. Here a solution involving the bad points X and Y: As |AB| = |BY|, one has that [angle AZB] = [angle ACB] = [angle BZY] = [angle BCY]. In the circle with center at B and passing through A, X and Y, one has that [angle AXY] + [angle ABY]/2 = 180°, which means that [angle AXZ] = [angle ABY]/2. As the points B and Z lie on opposite sides of AY, one has that [angle ABY] + [angle AZY] = 180° = 2*[angle AXZ] + 2*[angle BZY], i.e. [angle AXZ] + [angle BZY] = 90°, which means that AC and BZ are perpendicular.
Nice! Indeed there are many ways to tackle this problem regardless of considering angles with bad points. However, this strategy is affective when handling angles in harder problems. I will try to post a video where I illustrate the whole principle with a really complicated angle chasing problem.
Another solution: B is on perp bisector of AY, so on angle bisector of AZY. So B is on angle bisector of AZX and on perp bisector of AX. But obviously ABXZ aren’t concyclic since X!=C. This means that the angle bisector of AZX is the same line as the perp bisector of AX. So Z is on perp bisector of AX. Since B is also onnthat perp bisector we have BZ perpendicular to AC
Well I came up with a difference solution using both "bad points". So let D be the intersection of |AY| and |BZ|. Notice that since |BY|=|BA| we have that B is the mid point of the arc AY so BZ is the angle bisector of [angle AZY] and used this with the fact that ABCZY is cyclic we'll have [angle AZB]=[angle YZB]=[angle BYA]=[angle BAY]. Now notice that ∆BYD~∆BZY because [angle BYD]=[angle BZY] and they have a common angle at B therefore [angle BDY]=[angle BYZ] but the ∆BXY is isosceles so [angle BYZ]=[angle BYX]=[angle BXY]=[angle BDY] so the quadrilateral BYXD is cylic thefore [angle DYX]=[angle XBD]. With this we have [angle AYZ]=[angle ABZ]=[angle XBD] so BZ is the angle bisector of [angle XBA ] and since ∆BXA is isosceles it is also it's perpendicular bisector so that finishes the problem.
Great job! To be honest, this problem is not that hard, that whatever angle you pick you can find a way to handle it. The strategy of reducing the number of bad points however becomes very important in much harder problems. I will upload a video explaining more with a much more suitable example.
Solution 3: Let H be the orthocentre of ABC. Then HBX=HBA=HCA=HCX, so HBCX concyclic. This gives BXC=BHC=180-BAC=180-BZC=BYC. Also XCB=YCB and XB=YB, so triangles XCB and YCB are congruent. That gives BC perpendicular to XY, so BC perpendicular to ZX. Combining this with BXC=180-BZC implies that X is orthocenter of BCZ, so BZ perpendicular to CX
After BXHC concyclic we can also prove BC perpendicular to XY by using that reflecting (BHC) in BC gives (BAC), so reflection X’ of X in BC is the point in (ABC) with BX’=BX=BA which is Y. So XY perpendicular to BC
Hey sir, I was wondering what I should do in order to prepare for math Olympiad. Im currently year 10 (gcse) and can u please recommend a few textbooks for the preparation.
Thanks for the video. Here a solution involving the bad points X and Y: As |AB| = |BY|, one has that [angle AZB] = [angle ACB] = [angle BZY] = [angle BCY]. In the circle with center at B and passing through A, X and Y, one has that [angle AXY] + [angle ABY]/2 = 180°, which means that [angle AXZ] = [angle ABY]/2. As the points B and Z lie on opposite sides of AY, one has that [angle ABY] + [angle AZY] = 180° = 2*[angle AXZ] + 2*[angle BZY], i.e. [angle AXZ] + [angle BZY] = 90°, which means that AC and BZ are perpendicular.
Nice!
Indeed there are many ways to tackle this problem regardless of considering angles with bad points. However, this strategy is affective when handling angles in harder problems.
I will try to post a video where I illustrate the whole principle with a really complicated angle chasing problem.
Nice solution
@@littlefermatPlease do it :).
Another solution: B is on perp bisector of AY, so on angle bisector of AZY. So B is on angle bisector of AZX and on perp bisector of AX. But obviously ABXZ aren’t concyclic since X!=C. This means that the angle bisector of AZX is the same line as the perp bisector of AX. So Z is on perp bisector of AX. Since B is also onnthat perp bisector we have BZ perpendicular to AC
Well I came up with a difference solution using both "bad points". So let D be the intersection of |AY| and |BZ|. Notice that since |BY|=|BA| we have that B is the mid point of the arc AY so BZ is the angle bisector of [angle AZY] and used this with the fact that ABCZY is cyclic we'll have [angle AZB]=[angle YZB]=[angle BYA]=[angle BAY].
Now notice that ∆BYD~∆BZY because [angle BYD]=[angle BZY] and they have a common angle at B therefore [angle BDY]=[angle BYZ] but the ∆BXY is isosceles so [angle BYZ]=[angle BYX]=[angle BXY]=[angle BDY] so the quadrilateral BYXD is cylic thefore [angle DYX]=[angle XBD].
With this we have [angle AYZ]=[angle ABZ]=[angle XBD] so BZ is the angle bisector of [angle XBA ] and since ∆BXA is isosceles it is also it's perpendicular bisector so that finishes the problem.
Great job!
To be honest, this problem is not that hard, that whatever angle you pick you can find a way to handle it.
The strategy of reducing the number of bad points however becomes very important in much harder problems.
I will upload a video explaining more with a much more suitable example.
@@littlefermatlooking forward to it.
@@littlefermatthanks
Solved in 1 minute, solution: angle chase BZX+ZXA=BZY+180-AXY=BAY+ABY/2=90, so BZ is perpendicular to AC
Solution 3: Let H be the orthocentre of ABC. Then HBX=HBA=HCA=HCX, so HBCX concyclic. This gives BXC=BHC=180-BAC=180-BZC=BYC. Also XCB=YCB and XB=YB, so triangles XCB and YCB are congruent. That gives BC perpendicular to XY, so BC perpendicular to ZX. Combining this with BXC=180-BZC implies that X is orthocenter of BCZ, so BZ perpendicular to CX
After BXHC concyclic we can also prove BC perpendicular to XY by using that reflecting (BHC) in BC gives (BAC), so reflection X’ of X in BC is the point in (ABC) with BX’=BX=BA which is Y. So XY perpendicular to BC
Beautiful solution
Hey sir, I was wondering what I should do in order to prepare for math Olympiad. Im currently year 10 (gcse) and can u please recommend a few textbooks for the preparation.
hi sir please can you solve this functional equation i try for more times but i can't solve it f(xf(y)-f(x))=2f(x)+xy