July is another "busy farming season." It is time to transplant rice seedlings in the countryside, sowing the seedlings and waiting for the next harvest season.
I have always felt that my mathematical ability leans toward analysis and calculation rather than geometry. Even when I encounter geometry problems, my mind is filled with analytic geometry approaches, lacking the beauty of pure geometry. Over the past few days, in order to strengthen my ability to solve mathematical competition problems, I have been looking at IMO problems and attempting to solve some of them independently, but to no avail. I am particularly interested in inequalities; I feel that a simple expression that can be explained clearly without too many words is the hallmark of an inequality problem. However, the IMO inequality problems are truly profound, and I have not yet been able to solve one independently (I can understand the reference solutions, but I just cannot think of the approach). Perhaps it is because I am striving for unified methods and am unwilling to study those specific tricks. Unexpectedly, I looked at a geometry problem from the 2001 IMO today and realized I might be able to solve it. After studying it for a while, I luckily managed to work it out. Although it is not the simplest method, I would like to share it with everyone.
As shown in the figure, O is the circumcenter of acute triangle ABC, AP is the altitude of the triangle, and \angle B - \angle C \ge 30^\circ. Prove that \angle BAC + \angle BOP < 90^\circ.
Actually, the reason BoJone could solve this problem is that it is truly not that difficult. My geometry level remains at the junior high school stage, and my proof still focuses on analysis and calculation. First, from \angle B - \angle C \ge 30^\circ, we can deduce that \angle B > 60^\circ and \angle A + \angle C < 120^\circ. Otherwise, \angle C would be a right or obtuse angle, which contradicts the problem statement.
Construct auxiliary lines as shown in the figure below: let OD be perpendicular to BC, and connect O and C. The three angles of the triangle are denoted directly as A, B, C.
Since the central angle is twice the corresponding inscribed angle, we have \angle BOC = 2A. Thus, \angle A = \angle BOD. The problem asks to prove \angle A + \angle BOP < 90^\circ, which is equivalent to proving \angle BOD + \angle BOP < 90^\circ. Since \angle BOD + \angle OBD = 90^\circ, this is equivalent to proving \angle BOP < \angle OBD, which in turn is equivalent to proving BP < OP, or BP^2 < OP^2.
Let the sides opposite to angles A, B, C be a, b, c respectively, and let the circumradius be R. It is easy to see that: BP = c \cdot \cos B By the Law of Cosines: \begin{aligned} OP^2 &= BP^2 + R^2 - 2R \cdot BP \cdot \cos \angle OBD \\ &= BP^2 + R^2 - 2R \cdot BP \cdot \sin \angle BOD \\ &= BP^2 + R^2 - 2R \cdot c \cdot \cos B \cdot \sin A \end{aligned}
Thus, the conclusion to be proved becomes: R^2 - 2R \cdot c \cdot \cos B \cdot \sin A > 0. From the Law of Sines, we know that 2R \cdot \sin C = c. Therefore, we ultimately need to prove: 4 \sin A \cdot \sin C \cdot \cos B < 1
Using the product-to-sum formula: \sin C \cos B = \frac{1}{2} [\sin(B+C) - \sin(B-C)]. Since B - C \ge 30^\circ, we have \sin(B-C) \ge 1/2.
Therefore: \begin{aligned} 4 \sin A \cdot \sin C \cdot \cos B &= 2 \sin A [\sin(B+C) - \sin(B-C)] \\ &< 2 \sin A (\sin A - 1/2) \\ &\leq 2 \cdot 1 \cdot (1 - 1/2) = 1 \end{aligned}
The proof is complete.
The proof method in this article belongs to the category of "using a sledgehammer to crack a nut"... it is just that BoJone’s geometry skills are really quite weak, and I could not think of a better proof method...
When reposting, please include the original link: https://kexue.fm/archives/1453
For more details on reposting, please refer to: Scientific Space FAQ