<div class='user_solution'><p><strong>1.Rh1+!</strong> (Try: <u>1.Rf1+? c1Q 2.Rxc1+ Kxc1 3.Kxd4 Kb2!</u> (not 3...Kc2? 4.Re7! zz,btm 4...Kb3 5.Kc5! Kc3 6.Re3+! Kd2 7.Ra3 Ke2 8.Kd6 Kd2 9.Ke7 Ke2 10.d6 Kd2 11.Kd8 Ke2 12.Kc8 +-) <u>4.Re7 </u>(4.Kc5 d6+! =; 4.Kc5 Ka3=)<u> 4...Kc2! zz,wtm 5.Kc5</u> (5.Kc4 Rc7+!=; 5.d6 Kb3 6.Kc5 Ka4 =) <u>5...Kd3! 6.d6</u> (6.Kd6 Kd4=) <u>6...Kc3! 7.Re2 Kd3 8.Re7 Kc3 9.Re8 Kd3</u> - positional draw.) <strong>1...c1Q 2.Rxc1+ Kxc1 3.Kxd4! </strong>(3.Kd3? Kb2! 4.Kd4 d6! =) <strong>3...d6</strong> (3...Kb2 4.Kc5 +-) <strong>4.Rf6!</strong> (Try: 4.Rd8? Rb6 5.Ke4 Kc2(b2,d2) 6.Kf5 Kd3(c3,b3) 7.Ke6 Kc4! 8.Rxd6 Rb8! 9.Rc6+ Kb5 10.Rc1 Rh8! 11.d6 Kb6! 12.d7 Rh6+ 13.Ke7 Rh7+ 14.Ke8 Rh8+ 15.Kf7 Rh7+ - positional draw) <strong>4...Rb6!</strong> (4...Rd7 5.Ke4 Kc2 6.Kf5! Kd3 7.Ke6 +-) <strong>5.Kc3!</strong> (5.Ke4? Kc2(d2) 6.Kf5 Kd3(c3) =) <strong>5...Kd1</strong> (5...Kb1 6.Rf8(f4,f7) Ka2 7.Ra4(a3) Kc1 8.Ra1+ Rb1 9.Rxb1+ +-) <strong>6.Re6!</strong> <strong>zz,btm 6...Re6 7.Kb4</strong>! (7.Kc4? Rb6 8.Kc3 Ra6 - loss of time; 7.Kd3? Kc1 8.Kc3 Kd1 - loss of time) <strong>7...Kc2 8.Kb5 Ra8 9.Rxd6</strong>, win.</p><p>A synthesis of different mutual zugzwangs in try and solution.</p></div>