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The innovative application of Polya's problem-solving thought 
in the teaching of mathematical problem-solving--Take a conic 

section of the college entrance examination as an example 
Yuwei Wu* 

Guangzhou University, School of Mathematics and Information Science, Guangdong, China 
*Corresponding author: vj162774@163.com 

Abstract. In view of the current situation of mathematics teaching thinking and the problems, which 
are easily overlooked in high school classrooms using Polya's problem-solving ideas. This paper 
innovates the "'how to solve it' list" in combination with the characteristics of the current era, which 
extends the "four steps to solve problems" into five steps. To take a conic section of the college 
entrance examination as the carrier, the five steps are applied in teaching mathematical problem-
solving. It is expounded in detail to provide a method reference for optimizing the teaching thinking 
of high school mathematics teachers. 

Keywords: Problem-solving Thinking; Teaching Thinking; Conic Section. 

1. The question is raised 

1.1 Mathematical Teaching Thinking and Its Present Situation 
Mathematics teaching thinking is the rational understanding of mathematics teaching activities 

(processes) by mathematics teachers, the internal rational activities of the interaction between 
mathematics teachers and mathematics teaching objects (textbooks, students, etc.) ability. 
Mathematics teaching thinking "focuses on teaching thinking that meets the needs of students, rather 
than thinking only for teachers themselves or the content of the subject."[1-3] 

The teaching thinking of mathematics teachers has always been an essential part of mathematics 
teaching, and its differences will directly affect the quality of teaching. While it is widely believed 
that mathematics is the gymnastics of the mind, its function as "mind gymnastics" has not been fully 
utilized. It is mainly due to the following three deficiencies in mathematics teaching. Firstly, it focuses 
on knowledge and skills while ignoring mathematical thinking. Secondly, it focuses on thinking 
results and neglects the thinking process [4-5]. 

1.2 Polya's problem-solving ideas and new mathematical problems 
In the book "How To Solve It," George Polya proposed four steps for solving problems in the 

process of thinking in mathematics, including "understanding the problem," "devising a plan," 
"carrying out the plan," and "looking back" four steps [6]."Understanding the problem" is to let 
students understand the meaning of the problem, and to understand the known and unknown 
quantities of the problem, which is the basis for solving the problem. "Devising a plan" is the core of 
problem-solving. "Carrying out the plan" is mainly to comprehensively express the previous step to 
ensure the accuracy of the results, which is the key to problem-solving." Looking back" is a "re-
understanding" of problem-solving activities and trying to draw inferences about the problem from 
one case to another, and then make logical presuppositions for solving this type of problem, which is 
the sublimation of problem-solving. 

Because most of the current problem-solving process and teaching focus on how to find the ideas 
and methods to solve the problem, teachers often ignore the "looking back" step in the teaching 
process. Students only know the problem-solving tactics and feel that the problem-solving process is 
rushed. They do not know how the solution was discovered, nor the extensive connection between 
the problem and other concepts, theories, methods, and problems, let alone how the problem should 
be solved and how the topic will develop. Even if there is a "looking back" step in the class, it is 



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limited to answering the question itself and to the process of "checking whether the answer is correct" 
or recalling the question answered in memory. Such a "looking back" is too formal and low-level, 
which cannot achieve the purpose of cultivating students' divergent thinking and critical thinking, and 
does not meet the true meaning of "looking back" originally referred to by Polya [7]. The reason is 
that teachers' thinking in the teaching process is too limited, and they blindly pursue how to "instill" 
success without showing a complete thinking process, and they lack a deep understanding of the 
meaning of review after solving problems. Therefore, it is necessary to discuss further the step of 
"looking back" to innovate the problem-solving thought of Polya. Therefore, the teaching thinking of 
mathematics teachers can be effectively improved, and the effect of mathematics teaching can be 
improved. 

2. The innovative thinking of Polya's problem-solving ideas 
In view of the existing problems and defects, this paper attempts to innovate Polya's four steps to 

solve problems. It proposes to divide them into five steps: "observing and doubting," "multivariate 
analyzing doubts," "accurately solving doubts," "reviewing and doubting," and "reflecting and 
doubting." "Looking back" is refined into two steps "reviewing and doubting" and "reflecting and 
doubting." 

2.1 Observing and doubting - finding the crux of the problem 
From observing and doubting, based on existing knowledge, it is conducted an in-depth analysis 

of doubts (topics), which clarifies the mathematical problems that need to be solved in the doubts. It 
is known that knowledge points are related to the problems and understanding the known conditions 
and unknowns of the topic. Moreover, it looks for the obstacles that need to be overcome from the 
condition to the conclusion to find the crux of the problem. 

2.2 Multivariate analyzing doubts - exploring the path of solving the problem 
Multivariate analyzing doubts, that is, multi-level and multi-angle analysis of doubts, aims to 

resolve the crux of the problem and find an effective solution. Recall whether there is a connection 
between the current problem and similar problems or similar models. If so, try to solve the problem 
in the same way. If not, it analyzes the relationship between the known conditions and the problem 
and thinks about what information can be obtained from the known and which information is related 
to the problem, or what conditions can be inversely deduced from the problem, and how the inversely 
deduced conditions are related to the known conditions. It is used to build a "bridge" between the 
known conditions and the problem and find the optimal solution path. 

2.3 Accurately solving doubts - accurately implementing the plan 
Accurately solving doubts means implementing the plan strictly and accurately to solve doubts, 

ensuring the accuracy of the problem-solving results, and aiming to test whether the formulated plan 
is feasible to achieve the purpose of problem-solving. "Accurately solving doubts" is much easier 
than "Multivariate analyzing doubts." It mainly implements the already formed problem-solving ideas 
on the test paper. The most important thing at this stage is to be patient and careful. In order to ensure 
that every detail is accurate, every step must be carefully checked. Every point should be clarified so 
that the entire problem-solving process is free from the ambiguity that may hide errors. 

2.4 Reviewing and doubting - perceive the problem-solving process 
Reviewing and doubting, that is, review and sort out the entire problem-solving process, reflect on 

whether the problem-solving process has found the right direction, whether there are loopholes in 
one's subject knowledge, and perceive the mathematical knowledge, mathematical methods, and 
mathematical ideas used in solving problems, and think about it. It can use different methods to solve 



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the problem and consider whether the problem can be summarized and classified to achieve the 
problem-solving effect of "doing one, knowing one class." 

2.5 Reflecting and doubting - the problem of sublimation and expansion 
Reflecting and doubting mean necessary extension and expansion of the original topic, reflecting 

on whether the problem-solving process can withstand scrutiny, and then discovering the rules. 
Reflecting and doubting are different from reviewing and doubting. It is the growth of problem-
solving thinking based on reviewing and doubting. As Polya said, "Mathematics is not a pile of 
symbols and formulas, but a lively intellectual activity." [8] This developmental teaching link 
stimulates students' subjective enthusiasm for learning and improves their understanding. It has 
accumulated valuable experience and laid a solid foundation for solving more, deeper, and wider 
mathematical problems in the future. 

3. Practical applications 
The topic (2021 National Unified College Entrance Examination Paper Science Volume B 

Question 21) has a known focus 𝐹𝐹 on the parabola 𝐶𝐶: 𝑥𝑥2 = 2𝑝𝑝𝑝𝑝(𝑝𝑝 > 0), and the minimum distance 
between 𝐹𝐹 and the point on the circle 𝑀𝑀: 𝑥𝑥2 + (𝑝𝑝 + 4)2 = 1 is 4. 

(1) Ask 𝑝𝑝. 
(2) If the point 𝑃𝑃 is on 𝑀𝑀, 𝑃𝑃𝑃𝑃 and 𝑃𝑃𝑃𝑃 are the two tangent lines of 𝐶𝐶 , 𝑃𝑃, and 𝑃𝑃 are the 

tangent points, find the maximum value of the area of ∆𝑃𝑃𝑃𝑃𝑃𝑃. 
In high school teaching, most students have reached the level of solving the first question, so only 

the second question is analyzed to illustrate further the innovative application value of Polya's 
problem-solving thought. 

In this paper, simulation teaching is carried out through dialogue. The code name of the teacher is 
"T," and the code name of the student is "S." [9-10] 

Analysis of the second question: 
(1) Observing and doubting 
T: Read the second question carefully and talk about what information you have obtained? 

Moreover, what is the unknown quantity of the title? Is this a question of what? 
S:𝑃𝑃𝑃𝑃 and 𝑃𝑃𝑃𝑃 are the two tangent lines of 𝐶𝐶, 𝑃𝑃, and 𝑃𝑃 are the tangent points. The unknown 

quantity is the maximum value of the area of ∆𝑃𝑃𝑃𝑃𝑃𝑃. It is a question of the best value. 
T: What is the most common method among the solutions to the conic section maximum value 

problem? 
S: Transform the conic section maximum value problem into the maximum value problem of 

quadratic function or trigonometric function, and then use the basic inequality method, the matching 
method, element replacement method, derivative method, number-shape combination method, and 
other methods to find the maximum value. 

T: In order to use these methods, what are the prerequisites? 
S: Convert ∆𝑃𝑃𝑃𝑃𝑃𝑃 area expressions into quadratic or trigonometric expressions. 
(2) Multivariate analyzing doubts 
At this time, students may be limited to finding connections between general conditions and 

∆𝑃𝑃𝑃𝑃𝑃𝑃 area expressions, so a teacher can guide students to use reverse thinking to start from the 
problem to find the relationship between known conditions and known conditions. 

T: How can I find the area of ∆𝑃𝑃𝑃𝑃𝑃𝑃? Have you seen similar problems before? 
S: Using the formula 𝑆𝑆∆𝑃𝑃𝑃𝑃𝑃𝑃 = 1

2
|𝑃𝑃𝑃𝑃| ∙ 𝑑𝑑, 𝑑𝑑 is the distance from a point 𝑃𝑃 to a straight line 𝑃𝑃𝑃𝑃. 

The conic section is often asked for |𝑃𝑃𝑃𝑃|, I remember its expression is |𝑃𝑃𝑃𝑃| = √1 + 𝑘𝑘2|𝑥𝑥1 − 𝑥𝑥2|, 
and 𝑑𝑑 can be expressed by directly applying the point-to-distance formula. 

T: What do you need to know to get further expressions? Can these expressions be obtained based 
on the existing conditions? 



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S: We should know the equation for straight-line 𝑃𝑃𝑃𝑃, and expressions for 𝑥𝑥1 + 𝑥𝑥2 and 𝑥𝑥1𝑥𝑥2. It 
is known that 𝑃𝑃𝑃𝑃 and 𝑃𝑃𝑃𝑃 are the two tangents of 𝐶𝐶, and 𝑃𝑃 and 𝑃𝑃 are the tangent points. Suppose 
the coordinates of points 𝑃𝑃 and 𝑃𝑃 are(𝑥𝑥1, 𝑝𝑝1), (𝑥𝑥2, 𝑝𝑝2). According to the knowledge of derivatives, 
𝑘𝑘𝑃𝑃𝑃𝑃 = 1

2
𝑥𝑥1,𝑘𝑘𝑃𝑃𝑃𝑃 = 1

2
𝑥𝑥2, then the equation of the straight line 𝑃𝑃𝑃𝑃 is 𝑝𝑝 − 𝑝𝑝1 = 1

2
𝑥𝑥1(𝑥𝑥 − 𝑥𝑥1), and the 

equation of the straight line 𝑃𝑃𝑃𝑃 is 𝑝𝑝 − 𝑝𝑝2 = 1
2
𝑥𝑥2(𝑥𝑥 − 𝑥𝑥2). 

T: Can you simplify the equation? Which hidden condition is not used? 
S: The two points 𝑃𝑃,𝑃𝑃 are on the parabola 𝐶𝐶:𝑝𝑝 = 1

4
𝑥𝑥2, and the two straight-line equations can 

be simplified as 𝑙𝑙𝑃𝑃𝑃𝑃:𝑝𝑝 = 1
2
𝑥𝑥1𝑥𝑥 − 𝑝𝑝1, 𝑙𝑙𝑃𝑃𝑃𝑃:𝑝𝑝 = 1

2
𝑥𝑥2𝑥𝑥 − 𝑝𝑝2. 

T: Can you find the equation of straight line 𝑃𝑃𝑃𝑃 if you know the equation of straight line 𝑃𝑃𝑃𝑃 
and straight-line 𝑃𝑃𝑃𝑃?  

S1: It is known that 𝑃𝑃 is the intersection of the straight line 𝑃𝑃𝑃𝑃 and the straight line 𝑃𝑃𝑃𝑃. Let the 
coordinates of the point 𝑃𝑃 be (𝑥𝑥0, 𝑝𝑝0), and substitute it into the equation of the straight line 𝑃𝑃𝑃𝑃 and 

the straight line 𝑃𝑃𝑃𝑃 to get�
𝑝𝑝0 = 1

2
𝑥𝑥1𝑥𝑥0 − 𝑝𝑝1

𝑝𝑝0 = 1
2
𝑥𝑥2𝑥𝑥0 − 𝑝𝑝2

, so the equation of straight line 𝑃𝑃𝑃𝑃 is 𝑝𝑝0 = 1
2
𝑥𝑥0𝑥𝑥 −

𝑝𝑝, that is, 𝑝𝑝 = 1
2
𝑥𝑥0𝑥𝑥 − 𝑝𝑝0. 

T: Some students do not understand why they can directly write the equation of a straight line 𝑃𝑃𝑃𝑃. 
Please explain. 

S1: According to the knowledge of plane geometry, it is known that two points have only one 
straight line. That is, two points determine a straight line. Because 𝑃𝑃(𝑥𝑥1,𝑝𝑝1),𝑃𝑃(𝑥𝑥2,𝑝𝑝2) are two 
different points, and their coordinates satisfy the equation 𝑝𝑝 = 1

2
𝑥𝑥0𝑥𝑥 − 𝑝𝑝0, that is, the coordinates of 

𝑃𝑃 and 𝑃𝑃 points are the solution of the equation 𝑝𝑝 = 1
2
𝑥𝑥0𝑥𝑥 − 𝑝𝑝0, the equation of line 𝑃𝑃𝑃𝑃 is 𝑝𝑝 =

1
2
𝑥𝑥0𝑥𝑥 − 𝑝𝑝0.  

T: How to find the expression of 𝑥𝑥1 + 𝑥𝑥2 and 𝑥𝑥1𝑥𝑥2? 
S: The equation of the simultaneous line 𝑃𝑃𝑃𝑃 and the parabola 𝐶𝐶can be obtained according to 

Veda's theorem. 
(3) Accurately solving doubts 
Let 𝑃𝑃(𝑥𝑥1, 𝑝𝑝1),𝑃𝑃(𝑥𝑥2,𝑝𝑝2),𝑃𝑃(𝑥𝑥0, 𝑝𝑝0), there is 𝑥𝑥02 + (𝑝𝑝0 + 4)2 = 1.  
Given 𝐶𝐶:𝑝𝑝 = 1

4
𝑥𝑥2, derivation of the function gives 𝑝𝑝′ = 1

2
𝑥𝑥.  

∴ The equation of the line 𝑃𝑃𝑃𝑃 is 𝑝𝑝 − 𝑝𝑝1 = 1
2
𝑥𝑥1(𝑥𝑥 − 𝑥𝑥1), that is, 𝑝𝑝 = 1

2
𝑥𝑥1𝑥𝑥 − 𝑝𝑝1. 

Similarly, the equation of the straight line 𝑃𝑃𝑃𝑃 can be obtained as 𝑝𝑝 = 1
2
𝑥𝑥2𝑥𝑥 − 𝑝𝑝2. 

∵ Point 𝑃𝑃 is the common point of the two lines. There is �
𝑝𝑝0 = 1

2
𝑥𝑥1𝑥𝑥0 − 𝑝𝑝1

𝑝𝑝0 = 1
2
𝑥𝑥2𝑥𝑥0 − 𝑝𝑝2

. 

∴ The coordinates of the points 𝑃𝑃 and 𝑃𝑃 satisfy the equation 𝑝𝑝0 = 1
2
𝑥𝑥0𝑥𝑥 − 𝑝𝑝. 

∴ The equation of the straight line 𝑃𝑃𝑃𝑃 is 𝑝𝑝0 = 1
2
𝑥𝑥0𝑥𝑥 − 𝑝𝑝, that is, 𝑝𝑝 = 1

2
𝑥𝑥0𝑥𝑥 − 𝑝𝑝0. 

Combine �
𝑝𝑝 = 1

2
𝑥𝑥0𝑥𝑥 − 𝑝𝑝0

𝑝𝑝 = 1
4
𝑥𝑥2

, get 𝑥𝑥2 − 2𝑥𝑥0𝑥𝑥 + 4𝑝𝑝0 = 0, 

According to Veda's theorem, we can get 𝑥𝑥1 + 𝑥𝑥2 = 2𝑥𝑥0, 𝑥𝑥1𝑥𝑥2 = 4𝑝𝑝0. 
Therefore, 

|𝑃𝑃𝑃𝑃| = �1 + (1
2
𝑥𝑥0)2�(𝑥𝑥1 + 𝑥𝑥2)2 − 4𝑥𝑥1𝑥𝑥2 = �1 + (1

2
𝑥𝑥0)2�4𝑥𝑥02 − 16𝑝𝑝0 = �4 + 𝑥𝑥02�𝑥𝑥02 − 4𝑝𝑝0. 



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The distance from the 𝑑𝑑 = �𝑥𝑥02−4𝑦𝑦0�

�4+𝑥𝑥02
 From the point 𝑃𝑃 to the straight line 𝑃𝑃𝑃𝑃. 

∴  𝑆𝑆∆𝑃𝑃𝑃𝑃𝑃𝑃 = 1
2

|𝑃𝑃𝑃𝑃| ∙ 𝑑𝑑 = 1
2

|𝑥𝑥02 − 4𝑝𝑝0|�𝑥𝑥02 − 4𝑝𝑝0 = 1
2

(𝑥𝑥02 − 4𝑝𝑝0)
3
2. 

And 𝑥𝑥02 − 4𝑝𝑝0 = 1 − (𝑝𝑝0 + 4)2 − 4𝑝𝑝0 = −𝑝𝑝02 − 12𝑝𝑝0 − 15 = −(𝑝𝑝0 + 6)2 + 21. 
It is known that −5 ≤ 𝑝𝑝0 ≤ −3 , so when 𝑝𝑝0 = −5 , 𝑆𝑆∆𝑃𝑃𝑃𝑃𝑃𝑃  obtains the maximum value of  

1
2

× 20
3
2 = 20√5. 

(4) Reviewing and doubting 
The teacher organizes students to review the problem-solving method of this sub-question and 

asks S2 to review how this problem-solving idea was born. 
S2: I used reverse thinking to think about the problem you mentioned, guessing and assuming that 

the expression of 𝑆𝑆∆𝑃𝑃𝑃𝑃𝑃𝑃 can be written out, and then the maximum value can be obtained according 
to the properties of the function. And it is known that 𝑆𝑆∆𝑃𝑃𝑃𝑃𝑃𝑃 = 1

2
|𝑃𝑃𝑃𝑃| ∙ 𝑑𝑑, then the problem direction 

is transformed into an expression for finding |𝑃𝑃𝑃𝑃| and 𝑑𝑑.  
T: A "good thought" produces good results. During the inspection, I saw that S3 used different 

methods and obtained results. We asked him to share his ideas. 
S3: In daily problem solving, I am accustomed to directly setting the equation of the straight line 

𝑃𝑃𝑃𝑃. According to the meaning of the question, I know that the straight line 𝑃𝑃𝑃𝑃 and the parabola 
𝐶𝐶 must have two intersections, so set 𝑙𝑙𝑃𝑃𝑃𝑃:𝑝𝑝 = 𝑘𝑘𝑥𝑥 + 𝑏𝑏, according to the slope of the straight line 𝑃𝑃𝑃𝑃 
and the straight line 𝑃𝑃𝑃𝑃 can be solved. The coordinates of point 𝑃𝑃 is (𝑥𝑥1+𝑥𝑥2

2
, 𝑥𝑥1𝑥𝑥2

4
), and then the 

coordinates of point 𝑃𝑃 can be simplified to (2𝑘𝑘,−𝑏𝑏). Simultaneous equations, Veda's theorem, and 
other methods are shown as follows. It is similar to the above. The expression of 𝑆𝑆∆𝑃𝑃𝑃𝑃𝑃𝑃 is finally 
converted into an expression containing 𝑏𝑏. 

T: Very good! The above two students have made a brief exposition of their problem-solving ideas, 
I believe some should inspire everyone. What do you gain from solving this problem? 

S4: I feel that reverse thinking is a common way of thinking in problem-solving. This question is 
not limited to looking for a "bridge" between conditions and problems but inversely looking for a 
"bridge" between problems and conditions, which is the breakthrough point of this problem. 

S5: I feel that it is essential to transform your mind. This problem transforms the problem of the 
maximum value of the  ∆𝑃𝑃𝑃𝑃𝑃𝑃 area into a function to find the maximum value and converts it into 
the maximum value of the expression to find the ∆𝑃𝑃𝑃𝑃𝑃𝑃 area. Continuous transformation is the key 
to the success of the problem. 

(5) Reflecting and doubting 
T: Please review the problem-solving process again after completing the problem-solving process. 

Maybe you will generate some new ideas and gain some unexpected discoveries. 
S6: (raises his hand) In this question, the focus of the parabola and the center of the circle are both 

on the 𝑝𝑝axis. If they are replaced by both on the 𝑥𝑥axis, the method must be the same. 
T: If you change it according to the method you said, can a new topic be adapted so that the final 

result remains the same? 
S6: 



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T: Very good! Did any of the other students have any surprises? Try to adapt a question. 
S7: Replacing the circle 𝑀𝑀with a straight line can also be solved by using the method of this 

problem. 

T: Have you ever thought about extending and promoting the method of solving this problem? How 
did this idea come about? 

S8: I am thinking, is the solution of the area maximum value problem in the conic section 
transformed into the function maximum value problem? In the past problems of conic sections, I 
found that the area formula was often written 𝑆𝑆 = 1

2
|𝑃𝑃𝑃𝑃| ∙ 𝑑𝑑, and then the most value was calculated 

based on the knowledge of the function. 
T: That is right! In other words, it is aimed at solving the problem of the maximum value of the 

area in the conic curve, which is the problem of "the maximum value of the function." This will 
provide a solution path for subsequent model-like problems. 

4. Summary and Outlook 
This paper explores the problems existing in mathematics teaching thinking and the current 

problem-solving teaching using Polya's problem-solving thought and puts forward the innovative 
thinking of "whether it is possible to 're-create' the 'four steps of problem-solving." On the original 
basis, the step of "Looking back" is innovatively decomposed into two steps of "reviewing and 
doubting" and "reflecting and doubting," and the names of the remaining three steps are changed. 
Finally, taking the topic of the conic section of the college entrance examination as an example, this 
paper discusses the innovative application of Polya's problem-solving thought in mathematical 
problem-solving, which strengthens the process of students inferring others from one case. It provides 
a method reference for optimizing high school mathematics teachers' teaching thinking. 

Polya's problem-solving thought provides a method for students to do problems and for teachers 
to give lectures. Whether it is past, present, or future, research on this problem-solving thought has a 
particular value. With the changing times and the emergence of various problems in teaching, Polya's 
problem-solving thought will continue to evolve, radiate and generate new values in the future. 

References 
[1]  [ America] D. A. Grouse. Research Handbook of Mathematics Teaching and Learning [M]. Shanghai 

Education Press, 1999. 
[2]  Yang Qian. The Modern Enlightenment of Polya's Mathematical Education Theory[J]. Journal of 

Mathematical Education, 2002(02):18-20. 
[3]  Ren Zhanghui. Theory of Mathematical Thinking [M]. Nanning: Guangxi Education Press, 1990. 
[4]  Li Changguan. The way of thinking: the focus of mathematics teaching [J]. Mathematics Bulletin, 2022, 

61(01): 20-23. 
[5]  Zhang Jianyue. Establish curriculum awareness and implement core literacy [J]. Mathematics Bulletin, 

2016(5): 1-4,14 
[6]  G. Polya. How To Solve It [M]. Princeton University Press, 1973. 

It is known that the focus 𝐹𝐹 of the parabola 𝐶𝐶:𝑝𝑝2 = 2𝑝𝑝𝑥𝑥(𝑝𝑝 > 0), and the minimum 
distance between 𝐹𝐹 and the point on the circle 𝑀𝑀: (𝑥𝑥 + 4)2 + 𝑝𝑝2 = 1 is 4. If the point 𝑃𝑃 is on 
𝑀𝑀, 𝑃𝑃𝑃𝑃 and 𝑃𝑃𝑃𝑃 are the two tangents of 𝐶𝐶, and 𝑃𝑃 and 𝑃𝑃 are the tangent points, find the 
maximum value of the area of ∆𝑃𝑃𝑃𝑃𝑃𝑃. 

It is known that the focus 𝐹𝐹 of the parabola 𝐶𝐶: 𝑥𝑥2 = 2𝑝𝑝𝑝𝑝(𝑝𝑝 > 0), and the minimum 
distance between 𝐹𝐹 and the point on the line 𝑙𝑙: 3𝑥𝑥 − 4𝑝𝑝 − 16 = 0 is 4. If the point 𝑃𝑃 is on 𝑀𝑀, 
𝑃𝑃𝑃𝑃 and 𝑃𝑃𝑃𝑃 are the two tangents of 𝐶𝐶, and 𝑃𝑃 and 𝑃𝑃 are the tangent points, find the maximum 
value of the area of ∆𝑃𝑃𝑃𝑃𝑃𝑃. 



BCP Education & Psychology ERHSS 2022 
Volume 6 (2022)  
 

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[10]  Zhou Chenchen. Talking about the application of Polya's four-step problem-solving method in 
mathematical problem-solving—taking a conic section question in the college entrance examination as an 
example [J]. Mathematics Learning and Research, 2020(05):133-134. 


	1. The question is raised
	1.1 Mathematical Teaching Thinking and Its Present Situation
	1.2 Polya's problem-solving ideas and new mathematical problems

	2. The innovative thinking of Polya's problem-solving ideas
	2.1 Observing and doubting - finding the crux of the problem
	2.2 Multivariate analyzing doubts - exploring the path of solving the problem
	2.3 Accurately solving doubts - accurately implementing the plan
	2.4 Reviewing and doubting - perceive the problem-solving process
	2.5 Reflecting and doubting - the problem of sublimation and expansion

	3. Practical applications
	4. Summary and Outlook
	References