Xidian University 2026 Entrance Examination Math Solutions

Part 1: Fill-in-the-Blank Problems#

Problem 1#

Question: Among 12 uniformly distributed points on a circle, the number of obtuse triangles that can be formed by choosing any 3 points is $\underline{120}$.

An inscribed triangle is obtuse if and only if its three vertices lie within the same open semicircular arc; that is, the span of the three points is no more than 5 equal arc intervals.

Let the first vertex be $A$ (12 choices), and let the clockwise span be $k \in \{2, 3, 4, 5\}$, with the farthest endpoint denoted by $C$.

The intermediate vertex $B$ can be chosen in $k - 1$ ways between $A$ and $C$.

Therefore,

$$12 \times \sum_{k=2}^5 (k - 1) = 12 \times (1 + 2 + 3 + 4) = 12 \times 10 = \mathbf{120}$$

(Alternatively, use the complement method: the total number is $\binom{12}{3} = 220$, the number of right triangles is $6 \times 10 = 60$, and the number of acute triangles is $40$. Thus, the number of obtuse triangles is $220 - 60 - 40 = 120$.)

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Problem 2#

Question: Let $A = \{1, 2, 3, 4\}$ and $B = \{5, 6, 7\}$. If $f: A \to B$, the probability that $f$ is non-increasing is $\underline{\dfrac{5}{27}}$.

The total number of functions is the total number of mappings from $A$ to $B$:

$$|B|^{|A|} = 3^4 = 81$$

For $f$ to be non-increasing, we require

$$f(1) \ge f(2) \ge f(3) \ge f(4)$$

This is equivalent to choosing 4 elements from $B = \{5, 6, 7\}$ with repetition allowed and arranging them in non-increasing order:

$$\text{Number of admissible sequences} = \binom{4 + 3 - 1}{4} = \binom{6}{4} = 15$$

Hence,

$$P = \dfrac{15}{81} = \mathbf{\dfrac{5}{27}}$$

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Problem 3#

Question: Let $f(x) = \mathrm{e}^x + 4x$ and $g(x) = \ln x + 2x$. The number of lines tangent to both curves $y = f(x)$ and $y = g(x)$ is $\underline{1}$.

Let the tangency points be $(x_1, f(x_1))$ and $(x_2, g(x_2))$, where $x_2 > 0$.

Equal slopes give

$$f'(x_1) = g'(x_2) \implies \mathrm{e}^{x_1} + 4 = \dfrac{1}{x_2} + 2 \implies x_2 = \dfrac{1}{\mathrm{e}^{x_1} + 2}$$

Equal intercepts give

$$f(x_1) - x_1 f'(x_1) = g(x_2) - x_2 g'(x_2)$$

After simplification, we obtain the following equation in $x_1$:

$$h(x_1) = \mathrm{e}^{x_1}(1 - x_1) + \ln(\mathrm{e}^{x_1} + 2) + 1 = 0$$

Its derivative is

$$h'(x) = \mathrm{e}^x\left(\dfrac{1}{\mathrm{e}^x + 2} - x\right)$$

It is readily seen that $h'(x)$ has exactly one zero, $x_0$, in $(0, 1)$, so the function first increases and then decreases.

As $x \to -\infty$,

$$h(x) \to \ln 2 + 1 > 0$$

Moreover, $h(x_0) > 0$, while as $x \to +\infty$,

$$h(x) \to -\infty$$

Therefore, the equation has no solution on $(-\infty, x_0)$ and exactly one real root on $(x_0, +\infty)$, where $h$ is strictly decreasing. Thus, there is exactly one common tangent.

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Problem 4#

Question: Let $a_n$ be the integer closest to $\sqrt{n}$. The sum of the first 2026 terms of the sequence $\{a_n\}$ is $\underline{60810}$.

$$a_n = k \iff k - \dfrac{1}{2} < \sqrt{n} < k + \dfrac{1}{2} \iff k^2 - k + 1 \le n \le k^2 + k$$

The value $k$ therefore occurs exactly

$$(k^2 + k) - (k^2 - k) = 2k$$

times.

The total number of terms in the first $m$ groups is

$$\sum_{k=1}^m 2k = m(m+1)$$

Taking $m = 44$, the number of terms is

$$44 \times 45 = 1980$$

The sum of the first 44 groups is

$$\sum_{k=1}^{44} k \cdot 2k = 2\sum_{k=1}^{44} k^2 = 2 \times \frac{44 \times 45 \times 89}{6} = 58740$$

The number of remaining terms is

$$2026 - 1980 = 46$$

All these 46 terms have value 45. Hence,

$$S_{2026} = 58740 + 46 \times 45 = 58740 + 2070 = \mathbf{60810}$$

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Problem 5#

Question: Suppose $(x^2 - x + 1)^{2026} = a_0 + a_1 x + a_2 x^2 + \cdots + a_{2026}x^{2026}$. Then

$$2a_0 + 3a_1 + 4a_2 + \cdots + 2028a_{2026} = \underline{\qquad}$$

Let

$$P(x)=(x^2-x+1)^{2026}=a_0+a_1x+a_2x^2+\cdots+a_nx^n$$

For the expression to be evaluated,

$$S=2a_0+3a_1+4a_2+\cdots+(n+2)a_n$$

Term decomposition

That is,

$$S=\sum\limits_{k=0}^nka_k+2\sum\limits_{k=0}^na_k$$

Set $x=1$:

$$P(1)=a_0+a_1+a_2+\dots+a_n=1$$

Differentiate $P(x)$:

$$P'(x)=a_1+2a_2x+3a_3x^2+\dots+na_nx^{n-1}$$

Substituting $x=1$ gives

$$P'(1)=a_1+2a_2+3a_3+\dots+na_n=\sum\limits_{k=0}^nka_k$$

Also,

$$P'(x)=2026(x^2-x+1)^{2025}\cdot(2x-1)$$

Thus, at $x=1$,

$$P'(1)=2026$$

Therefore,

$$S=P'(1)+2P(1)=2026+2=2028$$

Constructive method

Because the coefficient index differs from the coefficient by 2, multiply both sides by $x^2$:

$$x^2P(x)=a_0x^2+a_1x^3+a_2x^4+\dots+a_nx^{n+1}$$

Differentiate the left-hand side:

$$\frac{\mathrm{d}[x^2P(x)]}{\mathrm{d}x}=2xP(x)+x^2P'(x)$$

Setting $x=1$ yields

$$S=2P(1)+P'(1)=2\times 1+2026=2028$$

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Problem 6#

Question: A rectangle has length 12 and width 8. If a semicircle can be placed entirely inside the rectangle, the maximum possible area of the semicircle is $\underline{(296 - 160\sqrt{3})\pi}$.

Oblique placement:

Place the semicircle at an angle such that its arc is tangent simultaneously to one long side and one short side of the rectangle, while the two endpoints of its diameter lie on the other two sides.

Suppose the center of the circle is $(r, r)$, and the two endpoints of the diameter lie on $x = m$ and $y = n$. The radius satisfies

$$(m-r)^2+(n-r)^2=r^2$$

or equivalently,

$$r^2-2(m+n)r+(m^2+n^2)=0$$

The maximum radius is therefore

$$r=m+n-\sqrt{2mn}=12+8-\sqrt{192}=20-8\sqrt{3}\approx 6.1436>6$$

The maximum area is

$$S=\dfrac{1}{2}\pi r^2=\mathbf{(296-160\sqrt{3})\pi}\approx18.87\pi\approx59.29$$

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Problem 7#

Question: Let $f(x)$ be a function defined on $\mathbf{R}$ satisfying $f(0)=0$, $f(x)+f(1-x)=1$, $f(x)=2f\left(\dfrac{x}{5}\right)$, and, whenever $0\le x_1\le x_2\le1$, $f(x_1)\le f(x_2)$. Then

$$f\left(\dfrac{1}{2026}\right)=\underline{\frac{1}{32}}$$

Since $f(0)=0$,

$$f(1)=1-f(0)=1$$

Then

$$f(1)=2f(1/5)\implies f(1/5)=1/2$$

By symmetry,

$$f(4/5)=1-f(1/5)=1/2$$

Combined with monotonicity, this shows that $f(x)$ is identically $1/2$ on $\left[\dfrac{1}{5},\dfrac{4}{5}\right]$ (a Cantor-like quinary staircase function).

By the recursive relation, when

$$x\in\left[\dfrac{1}{5^{k+1}},\dfrac{4}{5^{k+1}}\right]$$

we have

$$5^kx\in\left[\dfrac{1}{5},\dfrac{4}{5}\right]$$

and therefore

$$f(x)=\frac{1}{2^k}f(5^kx)=\frac{1}{2^{k+1}}$$

Take $k=4$. Then the interval is

$$\left[\dfrac{1}{3125},\dfrac{4}{3125}\right] = \left[\dfrac{1}{3125},\dfrac{1}{781.25}\right]$$

Since

$$781.25<2026<3125$$

we have

$$\frac{1}{2026}\in \left[\dfrac{1}{3125},\dfrac{4}{3125}\right]$$

Consequently,

$$f\left(\frac{1}{2026}\right) = \frac{1}{2^{4+1}} = \mathbf{\frac{1}{32}}$$

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Problem 9#

Question: Let $S=\{1,2,\cdots,2026\}$. The number of two-element sets $\{A,B\}$ satisfying $A,B\subseteq S$ and $A\cap B=\emptyset$ is

$$\underline{\dfrac{3^{2026}-1}{2}}$$

First consider ordered pairs $(A,B)$ satisfying $A\cap B=\emptyset$. Each element of $S$ has three possible assignments: it belongs to $A$, it belongs to $B$, or it belongs to neither. Thus, there are

$$3^{2026}$$

ordered pairs.

The problem requires $\{A,B\}$ to be a two-element set, meaning that it is unordered and its two elements are distinct, so $A\ne B$.

If $A=B$ and $A\cap B=\emptyset$, then

$$A=B=\emptyset$$

Thus, only the pair $(\emptyset,\emptyset)$ must be excluded.

After removing this identical case, divide by 2 to account for unordered pairs:

$$\dfrac{3^{2026}-1}{2}$$

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Problem 10#

Question: Given $\{\sin\theta,\sin 2\theta,\sin 3\theta\} = \{\cos\theta,\cos 2\theta,\cos 3\theta\}$, determine $\theta$.

The sums of the elements of the two sets are equal:

$$(\sin\theta+\sin3\theta)+\sin2\theta = (\cos\theta+\cos3\theta)+\cos2\theta$$

Using the sum-to-product identities,

$$2\sin2\theta\cos\theta+\sin2\theta = 2\cos2\theta\cos\theta+\cos2\theta$$

Therefore,

$$(2\cos\theta+1)(\sin2\theta-\\cos2\theta)=0$$

If

$$2\cos\theta+1=0$$

then $\cos\theta=-1/2$. In this case, the resulting collection contains repeated elements and therefore does not satisfy the definition of a set with distinct elements; this case is discarded.

If

$$\sin2\theta=\cos2\theta$$

then

$$\tan2\theta=1 \implies 2\theta=\dfrac{\pi}{4}+k\pi \implies \theta=\dfrac{\pi}{8}+\dfrac{k\pi}{2}$$

where $k\in\mathbb{Z}$.

Verification: when $\theta=\pi/8$, both sets are

$$\left\{\sin\dfrac{\pi}{8},\dfrac{\sqrt{2}}{2},\cos\dfrac{\pi}{8}\right\}$$

whose elements are distinct, so the two sets are indeed equal. Hence,

$$\boxed{\theta=\dfrac{\pi}{8}+\dfrac{k\pi}{2},\quad k\in\mathbb{Z}}$$

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Part 2: Analytical Problem#

Question: Solve the inequality

$$\dfrac{1}{\sqrt{x+1}} \le \dfrac{1}{3}\sqrt{\dfrac{x}{x-1}}$$

Domain:

• $x+1>0\implies x>-1$;
• $\dfrac{x}{x-1}\ge0\implies x\le0$ or $x>1$;
• Therefore, the domain is $(-1,0]\cup(1,+\\infty)$.

When $x=0$, the left-hand side is $1$ and the right-hand side is $0$, so $1\le0$ is false. Thus, $x=0$ is excluded.

When $x\in(-1,0)$, both sides are positive, so we may square both sides:

$$\dfrac{1}{x+1}\le\dfrac{x}{9(x-1)}$$

Since $x-1<0$, clearing denominators reverses the inequality:

$$9(x-1)\ge x(x+1) \implies x^2-8x+9\le0$$

The discriminant is

$$\Delta=28$$

and the two roots are

$$4\pm\sqrt7>1$$

so there is no solution in $(-1,0)$.

When $x\in(1,+\infty)$, clearing denominators preserves the direction of the inequality:

$$9(x-1)\le x(x+1) \implies x^2-8x+9\ge0$$

Thus,

$$x\le4-\sqrt7 \quad\text{or}\quad x\ge4+\sqrt7$$

Since

$$4-\sqrt7\approx1.354>1$$

combining this with the domain gives

$$\boxed{x\in(1,4-\sqrt7]\cup[4+\sqrt7,+\infty)}$$

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Part 3: Analytical Problem#

Question: Find the product of the distances from one vertex of a regular 2026-gon inscribed in the unit circle to the other 2025 vertices.

Place the regular $n$-gon on the unit circle in the complex plane. Its vertices are the roots of

$$z^n-1=0$$

namely,

$$\omega_k=\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n}} \quad (k=0,1,\dots,n-1)$$

Take the reference vertex to be $z=1$, corresponding to $k=0$. Its distances to the other vertices are $|1-\omega_k|$.

Using the algebraic identity

$$z^n-1 = (z-1)(z^{n-1}+z^{n-2}+\cdots+z+1) = (z-1)\prod_{k=1}^{n-1}(z-\omega_k)$$

for $z\ne1$, divide both sides by $z-1$:

$$\prod_{k=1}^{n-1}(z-\omega_k) = z^{n-1}+z^{n-2}+\cdots+z+1$$

Taking the limit as $z\to1$ and then taking absolute values on both sides gives

$$\prod_{k=1}^{n-1}|1-\omega_k| = |1+1+\cdots+1| = n$$

For $n=2026$, the required product is

$$\boxed{2026}$$

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Part 4: Analytical Problem#

Question: Let the sequence $\{x_n\}$ satisfy

$$\left|x_n-\dfrac{x_{n+1}}{2}\right|\le1$$

and

$$|x_n|\le\left(\dfrac{3}{2}\right)^n$$

Prove that

$$|x_n|\le2$$

By the triangle inequality,

$$|x_n|\le1+\dfrac{1}{2}|x_{n+1}|$$

Iterating this inequality backward $k$ times gives

$$|x_n| \le 1+\frac12+\frac14+\cdots+\frac{1}{2^{k-1}} +\frac{1}{2^k}|x_{n+k}| = 2\left(1-\frac{1}{2^k}\right) +\frac{1}{2^k}|x_{n+k}|$$

Using the given growth bound,

$$|x_{n+k}| \le \left(\dfrac32\right)^{n+k}$$

Therefore,

$$\frac{1}{2^k}|x_{n+k}| \le \frac{1}{2^k}\left(\frac32\right)^{n+k} = \left(\frac32\right)^n\left(\frac34\right)^k$$

Since the common ratio satisfies

$$\frac34<1$$

we have

$$\lim_{k\to\infty} \left(\dfrac32\right)^n\left(\dfrac34\right)^k = 0$$

For any fixed $n$, letting $k\to\infty$ yields

$$|x_n|\le2$$

as required.

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Part 5: Analytical Problem#

Question: Let $f(x)=|x^2-a|$, and let $M(a)$ be the maximum value of $f(x)$ on $[0,1]$. Find the minimum value of $M(a)$.

Set

$$g(x)=x^2-a$$

For $x\in[0,1]$, we have $x^2\in[0,1]$, so the range of $g(x)$ is $[-a,1-a]$.

The maximum of $f(x)=|g(x)|$ is attained at an endpoint of this interval:

$$M(a) = \max_{x\in[0,1]}|x^2-a| = \max\{|a|,|1-a|\}$$

By the triangle inequality for absolute values,

$$M(a) = \max\{|a|,|1-a|\} \ge \frac{|a|+|1-a|}{2} \ge \frac{|a+(1-a)|}{2} = \frac12$$

Equality holds if and only if

$$|a|=|1-a|$$

which gives

$$a=\dfrac12$$

Therefore, the minimum value of $M(a)$ is

$$\boxed{\dfrac12}$$

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Part 6: Analytical Problem#

Question: Points $A$, $B$, and $C$ lie on the curve $xy=1$, and $\triangle ABC$ is an isosceles right triangle. Find the minimum possible area of $\triangle ABC$.

Let the right-angle vertex be

$$A\left(u,\frac1u\right)$$

and let

$$B\left(v,\frac1v\right)$$

Then

$$\overrightarrow{AB} = \left(v-u,\dfrac1v-\\dfrac1u\right)$$

Rotating $\overrightarrow{AB}$ by $90^\circ$ gives the coordinates of $C$:

$$x_C = u-\left(\frac1v-\frac1u\right), \qquad y_C = \frac1u+(v-u)$$

Since $C$ lies on the hyperbola $xy=1$, substituting $x_Cy_C=1$ and factoring yields the root $v=u$, which corresponds to a degenerate configuration. The other nondegenerate real root is

$$v=\frac{u^2-1}{u(u^2+1)}$$

The square of the leg length is

$$|\overrightarrow{AB}|^2 = (v-u)^2+\left(\frac1v-\frac1u\right)^2 = \frac{2(u^4+1)^3}{u^2(u^4-1)^2}$$

The area is

$$S=\dfrac12|\overrightarrow{AB}|^2$$

Let

$$t=u^2, \qquad w=t+\dfrac1t>2$$

After simplifying the area function, we obtain

$$S(w)=\frac{w^3}{w^2-4}, \qquad w>2$$

Differentiate to locate the extremum:

$$S'(w) = \frac{w^2(w^2-12)}{(w^2-4)^2} = 0 \implies w=\sqrt{12}=2\sqrt3$$

Substitution gives the minimum area:

$$S_{\min} = \frac{(2\sqrt3)^3}{(2\sqrt3)^2-4} = \frac{24\sqrt3}{8} = \boxed{3\sqrt3}$$