This paper has 19 problems, and we'll break them down one by one: first see how the original paper asks each question, then build the diagram step by step, and finally compare the key solution points.
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vizchat/gaokao-2026-quanguo-1-math · Published Oct 2026
Updated Oct 21, 2026
A question-by-question breakdown of 19 real exam problems: locating each item in the original paper, growing the diagrams step by step, and comparing key solution points.
This paper has 19 problems, and we'll break them down one by one: first see how the original paper asks each question, then build the diagram step by step, and finally compare the key solution points.
Question 1 is a gimme, but the point it gives away is “do you sort first?” Sorting isn't a problem-solving trick; it's written into the definition of the median—with the five numbers as given, the middle value is 4, and only after sorting is it 6.
Place the original order and the sorted order side by side. The position right in the middle is the answer.
After sorting: 4, 5, 6, 8, 12. With five numbers, take the third from the middle; the answer is 6, so choose B. This point isn't about calculation, but about the definition—taking the middle as-is gives 4, which is exactly the trap the problem setter left.
Question 2: a and b are not collinear, and 2a+yb = xa−3b. “Not collinear” is the key phrase—it makes “corresponding coefficients are equal” a valid deduction.
Draw a and b as two non-collinear vectors. If the left side 2a+yb and the right side xa−3b are to be the same vector, the two paths must land at the same point.
Return to that reasoning: only non-collinear vectors have a unique representation. Corresponding coefficients are equal, so x=2 and y=−3, choose A.
Question 3 tests special values of trigonometric functions, not set operations—calculate the three values, and the intersection is visible at a glance.
The three elements of A are calculated as −1/2, 1/2, 1. Put the elements of B in as well: the two shared elements automatically fall into the middle region.
The three values are −1/2, 1/2, and 1. Comparing with B, the intersection is −1/2 and 1, choose C.
Question 4: the tangent to y=5x+8ln x at (1,5). The derivative gives the slope, the point gives the location; with both pieces, you can write it.
First draw the curve; note that the domain is x>0. The tangent line is a derived object: given the curve and the point of tangency, it is computed on its own—change the point of tangency, and the tangent line follows.
The derivative at x=1 equals 13, which is the slope. Write it in point-slope form: the tangent line is y=13x−8, choose D.
Question 5: two parabolas with different opening directions both pass through (4,8). Solve for p in each, and the foci are determined.
Draw the two parabolas; one of their intersection points is exactly (4,8). The foci are derived—they follow the curves, not manually entered coordinates. Measure the distance between the two foci.
p₁=8 and p₂=1; the foci are (4,0) and (0,1/2), respectively. The distance is √65/2, choose D.
Question 6: f(x)=(x+2)/(eˣ+a) has maximum value 1; find a. The maximum-value condition gives two things at once—the derivative is zero, and the function value is 1.
a is a bound parameter—change it, and the whole curve deforms. Mark the horizontal line y=1 at the peak: only when a=1 does the curve just touch it without crossing above.
Combine “derivative is zero” with “function value is 1”; the denominator step forces e^{x₀} to become 1. Thus x₀=0 and a=1, choose B.
Question 7 has a real-world background, but the math has only two layers: first write out the pagoda counts for the 12 rows, then ask how to divide them into 6 groups so that the group sums form an arithmetic progression.
First lay out the 12 rows' pagoda counts according to the problem: the first five rows are given, and only from the sixth row onward do they form an arithmetic progression with common difference 2. Put the grouping scheme into a table—two numbers per group, and the group sums increase by 4 each time.
The total gives 2s₁+5d=36. d=4 can actually be matched, and the answer is 4.
Question 8: from 64 points, remove P(1,1,1), and find the expectation of x₁+x₂+x₃ over the remaining 63 points. Direct enumeration is a dead end; symmetry is the way in.
Each coordinate independently takes values −2, −1, 1, 2 with equal probability. The mean line falls at 0—this is why the sum over all of U is zero.
The sum over U is zero, and the removed point has X=3. The expectation is −1/21, choose A.
Question 9's four options each test one thing: the conjugate, the modulus, the square, and whether a quotient is real. Check them one by one; don't guess.
Plot z on the complex plane: the horizontal axis is the real part, the vertical axis the imaginary part. The conjugate is the mirror image across the real axis; squaring squares the modulus and doubles the argument, so z² moves farther out.
The modulus is √13, not 5, so B is wrong. The other three are correct; choose ACD.
Question 10: the distances from C and D to line AB are fixed, and the dihedral angle is fixed, but both can slide freely along the direction of AB. Two of the four options are broken by this degree of freedom.
AB is the edge, and C and D each lie in one of the half-planes. Draw the two perpendicular segments to the edge—lengths 2 and 1—and the angle between them is the 60° dihedral angle. Connect CD: its length varies only with the offset between the two feet of the perpendiculars, and it is shortest when the feet coincide.
The freedom to slide along the edge lets ∠CAD become arbitrarily small, so A is wrong; under the same freedom, AC·AD is always positive, so D is wrong. CD ⩾ √3 and “when AB⊥CD, CD⊥plane ABD” hold, so choose BC.
Question 11: three circles of radius 1 have centers forming an equilateral triangle of side length 2. A chord length is determined only by the distance from the center to the line, so the four options ask about different sides of the same thing.
The three centers form an equilateral triangle. Place a line through the origin, with the slope as a bound parameter. When the slope changes, the three chords change together—s₁+s₂+s₃ reaches its maximum at some slope.
k=±√3/3 are the only slopes that cannot be attained, so A is wrong. There are exactly 3 equidistant lines, infinitely many with sum 3, and when b=0 the maximum is 2√21/3; choose BCD.
Question 12: 5x²−6y²=1. The right side is 1 and nothing else, so a² and b² can be read off directly—first convert to standard form, then plug into the formula.
Draw the hyperbola. The asymptotes are derived objects—they are computed from the curve itself, and the slopes ±b/a are exactly what the ratio of a to b looks like on the graph.
a²=1/5, b²=1/6, and for a hyperbola c² is their sum. The eccentricity is √66/6.
Question 13: f(x)=2sin(ax+θ) is an even function and is increasing on (0, π/2). Evenness locks θ to two candidates, and monotonic increase chooses one of them.
Use π intervals on the horizontal axis. First draw 2cos x: it decreases on (0, π/2). Then draw −2cos x: it increases on the same interval. Both candidates for |a| pass through (2π/3, 1)—that is what it means for the answer to be independent of a.
Evenness implies θ can only be π/2 or 3π/2, and increasing picks out the latter. f(2π/3)=1, and both |a|=1 and 2 give the same value.
Question 14: the condition gives only S₃ₙ, meaning the sums of each three-term group are pinned down. We need to fit a geometric progression of 9 consecutive terms inside, and the common ratio can be at most what?
Difference S₃ₙ: the sum of each three-term group forms the arithmetic progression 2, 4, 6, 8. If 9 consecutive terms exactly cover three whole groups, the ratio of adjacent group sums would have to equal two different numbers at once—that is where the contradiction lies.
Covering three whole groups has no solution, so it must be offset. The two whole groups in the middle give q³=(m+2)/(m+1); it is largest when m=1, and q=∛(3/2).
Question 15, 13 points, tests line-plane parallelism and distance in a right triangular prism. There are three conditions: the base angle ∠ACB is a right angle, AC and BC are equal, and D and E are the midpoints of AB and AC₁, respectively. Note that the base is an isosceles right triangle, not an equilateral triangle—this determines how we draw it.
First draw the prism. The base is an isosceles right triangle, so here we use a free polyhedron: specify six vertices and five faces. Mark the two midpoints D and E, and draw DE. Then draw the perpendicular segment for “the distance from D to plane BCC₁B₁” as a dashed line—it lies inside the solid, exactly where a dashed line should appear.
Use the 45° condition to determine a. The answer is 1—when a line is parallel to a plane, the distance from the line to the plane is the distance from any point on the line to the plane, and taking D is the easiest.
Question 16 has two parts: first use the law of cosines to find cos A, then find CE under the constraints DE∥BC and AE⊥AC. The two conditions in the second part each control one direction.
First set up the isosceles triangle: AB=AC=3, BC=2√3. D lies on the extension of BA; through it draw a line parallel to BC—the parallel line is a derived object, depending on the point and the line it is parallel to. Then through A draw a line perpendicular to AC; its intersection with that parallel line is E.
cos A = 1/3, and the triangle is isosceles. The two conditions bind t and λ into t=2λ; DE=√6 determines λ, and finally CE=3√5.
Question 17, 15 points. Read the rule carefully: at most N shots are taken, and you stop if and only if you make one shot or miss all N. In other words, X is “the number of shots up to and including the first make,” but capped at N. This “cap” is the key to the next two parts.
Draw the probability distribution for N=4, p=1/3 as a chart. Note that the last bar is taller than the preceding pattern would suggest—that is exactly the merging caused by “whether or not the 4th shot goes in, you stop.” The expectation line is marked here; although this question doesn't ask for it, it makes the center of mass of the distribution immediately visible.
The second part uses the same observation: X greater than k means the first k shots all missed. Then the conditional probability simplifies to q to the m-th power—memorylessness.
Question 18, 17 points, is the final analytic geometry problem on this paper. The left focus is F(−1,0) and the eccentricity is 1/2; the first part asks for the equation. The two subparts of the second part both revolve around the same diagram: a moving line through F intersects the ellipse at P and Q, and line PO intersects the ellipse again at R. There is a key observation here—O is the origin, so R is the reflection of P across the origin.
First draw the ellipse and the left focus—the focus is a derived object, following the curve, not a manually entered coordinate. Then place a moving line through F: its slope is bound to parameter k, and as k changes, P and Q slide along the ellipse. P and Q are the two intersection points of the line and the ellipse, also derived; R is the reflection of P across the origin.
Fix the line using the answer to the first subpart: slope √5/2. At this point |y_Q| is exactly half of |y_P|, which is where the area relationship comes from. Connect triangle PQR and measure ∠PQR. The second subpart asks for the minimum tangent of this angle—convert tan into 3m + 4/m, and the AM-GM inequality gives 4√3.
Question 19 packages “whether the function becomes larger after a certain point” into a set D(x₀). The three parts build progressively; the first part makes this set concrete by calculation.
The branch for x<0 is 2ˣ, and the branch for x⩾0 is 1−x, starting at 1 at x=0. The horizontal line y=1/2 is at the height of f(−1): the segment where the function is above it, shifted back, is D(−1).
The two segments of D(−1) join exactly to form (0, 3/2). The common pivot for the last two parts: D translates size relationships into set inclusion, so monotonicity becomes a chain of inclusions.
The 19 problems are done. Looking back: which type of problem should use which kind of diagram is actually quite stable—for analytic geometry, draw curves and derived points; for solid geometry, build a polyhedron; for probability, draw a distribution; for sequences, look at the general term and the first n terms. Remembering this comparison table is more useful than remembering the answer to any one problem.
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