HMMT 二月 2024 · 团队赛 · 第 7 题
HMMT February 2024 — Team Round — Problem 7
题目详情
- [50] Let ABCDEF be a regular hexagon with P as a point in its interior. Prove that of the three values tan ∠ AP D , tan ∠ BP E , and tan ∠ CP F , two of them sum to the third one.
解析
- [50] Let ABCDEF be a regular hexagon with P as a point in its interior. Prove that of the three values tan ∠ AP D , tan ∠ BP E , and tan ∠ CP F , two of them sum to the third one. Proposed by: Albert Wang Solution 1: E D X P O F C Y Z Q A B WLOG let the side length of the hexagon be 1 . Let O be the center of the hexagon. Consider drawing in the circles ( AP D ) , ( BP E ) , and ( CP F ) . Note that O lies on the radical axis of all three circles, since AO · OD = BO · OE = CO · OF . Since P also lies on the radical axis, all three circles are coaxial. Let X , Y , and Z be the centers of ( AP D ) , ( BP E ) , and ( CP F ) , respectively. Since the circles are coaxial, X , Y , and Z are collinear. WLOG Y lies on segment XZ . Note that XO ⊥ AD, Y O ⊥ ◦ ◦ BE = ⇒ ∠ XOY = 60 . Similarly, we have ∠ Y OZ = 60 . Now, inverting at O and using van Schooten’s Theorem gives that 1 /OY = 1 /OX + 1 /OZ . Furthermore, we have 1 AO 1 ◦ ◦ ∠ AP D = 180 − ∠ AXD = 180 − ∠ AXO = ⇒ tan ∠ AP D = − tan ∠ AXO = − = − . 2 OX OX 1 1 Similarly, we have tan ∠ BP E = − and tan ∠ CP F = − . Therefore, two of these tangent values OY OZ sum to the third, as desired. Solution 2: Firstly, note that AD , BE , and CF are diameters of the circle ( ABCDEF ) , so the angles ∠ AP D , ∠ BP E , ∠ CP F are all obtuse. Therefore, the desired tangents are well-defined. WLOG let the side length of the hexagon be 1 . Let O be the center of the hexagon, and let OP = x . Finally, let ∠ AOP = θ . √ √ 2 2 Now, by Law of Cosines, we have AP = x + 1 − 2 x cos θ and DP = x + 1 + 2 x cos θ . Now, by Law of Cosines again we have 2 2 2( x + 1) − 4 x − 1 √ √ cos ∠ AP D = = 2 2 2 2 2 2 ( x + 1) − 4 x cos θ 2 2 2 (1 − x ) + 4 x sin θ 2 x | sin θ | = ⇒ tan ∠ AP D = . 2 x − 1 ◦ ◦ Similarly, ∠ ( BE, OP ) and ∠ ( CF, OP ) are θ +60 and θ +120 , respectively (here we are using directed angles). Therefore, the desired three values are { } ◦ ◦ 2 x | sin θ | 2 x | sin( θ + 60 ) | 2 x | sin( θ + 120 ) | { tan ∠ AP D, tan ∠ BP E, tan ∠ CP F } = , , . 2 2 2 x − 1 x − 1 x − 1 2 x ◦ ◦ We can scale down the tangent values by to get {| sin θ | , | sin( θ + 60 ) | , | sin( θ + 120 ) |} . Now, 2 x − 1 consider an equilateral triangle with vertices at the third roots of unity rotated by θ degrees counter- clockwise. The three values represent the distances from the three vertices to the real axis. Since the centroid of this triangle is the origin (lying on the real axis), two of these quantities must sum to the third, as desired. Solution 3: We will show either the three sum to 0 or two of them sum to the third one; since they’re all negative, the former case is actually impossible. πi/ 3 Let ( ABCDEF ) be the unit circle, with a = 1 , b = ω , and so on, where ω = e . Then ∠ AP D is the argument of 3 3 3 2 1 − p (1 − p )( ω − p ) ω − p − ω p + | p | = = . 3 3 3 2 3 3 ω − p ( ω − p )( ω − p ) 1 + | p | − ω p − ω p Then tan ∠ AP D is the imaginary part divided by the real part of this, which is 1 p − p − · = c · dist( P, AD ) 2 i | p | − 1 for some constant c . (Note that tan AP D might actually be the negative of this, depending on direction; this is why we added the remark at the beginning about them possibly summing to 0 .) Similarly, tan ∠ BP E = c · dist( P, BE ) and tan ∠ CP F = c · dist( P, CF ) . It suffices to show that two dist( P, AD ) , dist( P, BE ) , and dist( P, CF ) sum to the third. However, this is easy; without loss of generality let P be inside OAB , and let the hexagon have side length 1 . Then √ 3 dist( P, AD ) + dist( P, BE ) = − dist( P, AB ) = dist( P, CF ) , 2 as desired.