Let be an acute scalene triangle. Let and be two distinct interior points of the segment such that . Suppose that: and are the feet of the perpendiculars from from to the lines and respectively. and are the feet of the perpendiculars from to the lines and respectively.
Prove that and intersect on the line .
Six boys and six girls are participating at a tango course. They meet every evening for three weeks (a total of 21 times). Each evening, at least one boy-girl pair is selected to dance in front of the others. At the end of the three weeks, every boy-girl pair has been selected at least once. Prove that there exists a person who has been selected on at least 5 distinct evenings.
Note: a person can be selected twice on the same evening.
Let be a triangle with incenter , and let be a point on side . Points and are chosen on lines and respectively such that is a parallelogram. Points and are chosen on side such that and are the angle bisectors of angles and respectively. Let be the circle tangent to segment , the extension of past , and the extension of past . Prove that is tangent to the circumcircle of triangle .
For a positive integer , consider the fractions The product of these fractions equals , but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of for which this is possible and prove that you have found them all.
I think the answer is 4sqrt2 but I think there's a flaw in my solution(please point it out if possible):
It's pretty obvious that CA is perpendicular to AB. Let CH be length y, AB=CD=x and the radius of the circle be r. Connect C with the tangent from H and let the point be P.
First note that CAB is a right triangle so r^2+x^2=2y^2 (1)(because BC = sqrt(2)*CH)
Then note that CPH is a right triangle so r^2+4^2=y^2 ===> r^2+16=y^2 (2)
Now let the point where the tangent from D to the circle be point Q. DCQ is a right triangle. The hypotenuse is x and one leg with side length r. So we need to find x^2-r^2. We can do this from out systems. Plug in our equation for y^2 into equation (1). r^2+x^2 = 2r^2+32 or x^2-r^2=32.
Therefore, the tangent has length sqrt(32) or 4sqrt2