problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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Example 1 As shown in Figure 2, given that $P A$ and $P B$ are tangents to circle $\Gamma$ at points $A$ and $B$, and the secant $P C D$ intersects circle $\Gamma$ at points $C$ and $D$. Point $Q$ is on segment $C D$. If $\angle D A Q = \angle P B C$, then $\angle D B Q = \angle P A C$.
(2003, National High School Math... | Prove as shown in Figure 2, connect $A B$.
Notice that
$$
\begin{array}{l}
\angle D A Q=\angle P B C=\angle B A C, \\
\angle A D Q=\angle A D C=\angle A B C .
\end{array}
$$
Then $\triangle A D Q \sim \triangle A B C \Rightarrow \frac{A D}{D Q}=\frac{A B}{B C}$
$$
\begin{array}{l}
\Rightarrow A B \cdot D Q=A D \cdot B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,473 |
Example 2 As shown in Figure 3, from a point $P$ outside the circle $\Gamma$, draw two tangents $PA$ and $PB$ to the circle $\Gamma$, with $A$ and $B$ being the points of tangency.
Draw a secant line through $P$ that intersects the circle $\Gamma$ at points $C$ and $D$. Draw a line through the point of tangency $B$ p... | Proof As shown in Figure 3, connect $B C$, $B A$, and $B D$.
From the problem, we know
$$
\begin{array}{l}
\angle A E B=\angle E A P=\angle A B C \\
\Rightarrow \triangle A B E \backsim \triangle A C B \\
\Rightarrow \frac{A B}{B E}=\frac{A C}{C B} . \\
\text { Also, } \angle A B F=\angle B A P=\angle A D B \\
\Rightar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,474 |
Example 3 In the right triangle $ABC$, $AD$ is the altitude on the hypotenuse $BC$. The line through the incenter of $\triangle ABD$ and the incenter of $\triangle ACD$ intersects sides $AB$ and $AC$ at points $K$ and $L$, respectively. Let $S$ and $T$ be the areas of $\triangle ABC$ and $\triangle AKL$, respectively. ... | Proof As shown in Figure 5, connect $D O_{1}$, $D O_{2}$, and $A O_{1}$.
$$
\begin{array}{l}
\text { From } \triangle D O_{1} O_{2} \sim \triangle A B C \\
\Rightarrow \angle D O_{1} O_{2}=\angle B \\
\Rightarrow B, D, O_{1}, K \text { are concyclic } \\
\Rightarrow \angle A K O_{1}=\angle B D O_{1}=\angle A D O_{1} \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,475 |
Example 4 Given that $C D$ is the altitude on the hypotenuse $A B$ of $\mathrm{Rt} \triangle A B C$, $\odot O_{1}$ and $\odot O_{2}$ are the incircles of $\triangle A D C$ and $\triangle B D C$ respectively, and the other common external tangent of the two circles intersects $B C$ and $A C$ at points $P$ and $Q$ respec... | Prove: As shown in Figure 6, connect $D O_{1}$, $D O_{2}$, and $O_{1} O_{2}$. Let $P Q$ intersect $C D$ at point $E$, and connect $O_{1} E$ and $O_{2} E$.
From $\triangle D O_{1} O_{2} \sim \triangle C A B$, we get
$\angle D O_{1} O_{2}=\angle A=\angle D C B$.
Therefore, $\angle O_{1} D O_{2}+\angle O_{1} E O_{2}=90^{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,476 |
As shown in Figure 7, quadrilateral $ABCD$ is a circumscribed convex quadrilateral, with points of tangency $E, H, F, G$ on sides $AB, BC, CD,$ and $DA$ respectively. Then $AC, BD, EF,$ and $GH$ are concurrent.
This is known as Newton's Theorem. | Proof: Let $E F$ intersect $A C$ at point $P$. Then
$$
\begin{array}{l}
\frac{A P}{A E}=\frac{\sin \angle A E P}{\sin \angle A P E}=\frac{\sin \angle C F P}{\sin \angle C P F}=\frac{P C}{C F} \\
\Rightarrow \frac{A P}{P C}=\frac{A E}{C F} .
\end{array}
$$
Similarly, let $G H$ intersect $A C$ at $P^{\prime}, \frac{A P^... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,477 |
Example 5 Given $\triangle ABC$ with $AB > AC$, its incircle touches side $BC$ at point $E$. Connecting $AE$ intersects the incircle at point $D$ (different from point $E$). Take a point $F$ on segment $AE$ different from point $E$ such that $CE = CF$. Connect $CF$ and extend it to intersect $BD$ at point $G$. Prove: $... | Prove that, as shown in Figure 8, draw the tangent line $MN$ of the incircle through point $D$, intersecting $AB$, $AC$, and $BC$ at points $M$, $N$, and $K$ respectively.
From $\angle K D E=\angle A E K=\angle E F C \Rightarrow M K \parallel G C$. By Newton's theorem, $B N$, $C M$, and $A E$ are concurrent. Using Ceva... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,478 |
Example 6 Given an acute triangle $\triangle ABC$, construct $\odot O$ with $BC$ as its diameter. Draw two tangents from point $A$ to $\odot O$, with the points of tangency being $P$ and $Q$. Let $H$ be the orthocenter of $\triangle ABC$. Prove that $P$, $H$, and $Q$ are collinear. (1996, China Mathematical Olympiad) | Prove that connecting $A H$ and extending it to intersect side $B C$ at point $n$, with other auxiliary lines as shown in Figure 10.
it: Note that $A D$ is the solution.
$$
\begin{array}{l}
\angle A P D + \angle A Q I = 180^{\circ} . \\
X A P^{2} = A K \cdot A B = A H \cdot A D \\
\Rightarrow \triangle A P H \sim \tria... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,479 |
Example 3 As shown in Figure 1, in $\triangle A B C$, $A B=A C=$ $4, P$ is a point on side $B C$ different from points $B, C$. Then the value of $A P^{2}+B P \cdot P C$ is ( ).
(A) 16
(B) 20
(C) 25
(D) 30 | Solution 1 As shown in Figure 1, draw $A D \perp B C$.
Since $A B=A C$, we have $B D=D C$.
$$
\begin{array}{l}
\text { Therefore, } A P^{2}+B P \cdot P C \\
=A D^{2}+P D^{2}+(B D-P D)(D C+P D) \\
=A D^{2}+P D^{2}+B D^{2}-P D^{2} \\
=A B^{2}=16 .
\end{array}
$$
Solution 2 Let point $P$ be the midpoint $D$ of $B C$, at ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,481 |
As shown in Figure $12, A B$ is the diameter of the semicircle $\odot O$, and the radius of $\odot 0^{\prime}$ is $r$. $\odot 0^{\prime}$ is internally tangent to the semicircle $\odot O$ and tangent to $A B$ at point $D$. Then
$$
\frac{1}{r}=\frac{1}{A D}+\frac{1}{D B} \text {. }
$$ | Proof: Let $A D=a, D B=b(a>b)$. Then
$$
\begin{array}{l}
O A=O B=\frac{1}{2}(a+b), O D=\frac{1}{2}(a-b), \\
O O^{\prime}=\frac{1}{2}(a+b)-r .
\end{array}
$$
By $O O^{\prime 2}=O D^{2}+O^{\prime} I D^{2}$ $\Rightarrow\left[\frac{1}{2}(a+b)-r\right]^{2}=\left[\frac{1}{2}(a-b)\right]^{2}+r^{2}$ $\Rightarrow r=\frac{a b}{... | \frac{1}{r}=\frac{1}{A D}+\frac{1}{D B} | Geometry | proof | Yes | Yes | cn_contest | false | 724,482 |
Example 8 As shown in Figure $13, A B$ is the diameter of the semicircle $\odot \emptyset$. Point $C$ moves on the semicircle $\odot O$, and $C D \perp A B$ at point $D$. $\odot O_1$ is internally tangent to arc $\overparen{A C}$ and tangent to $A B, C D$. $\odot O_2$ is internally tangent to arc $\overparen{C B}$ and ... | Proof: Let $A D=a, D B=b, O_{2} F=x$. Then
$$
\begin{array}{l}
\frac{1}{O_{2} F}=\frac{1}{A F}+\frac{1}{B F} \\
\Rightarrow \frac{1}{x}=\frac{1}{a+x}+\frac{1}{b-x} \\
\Rightarrow x=\sqrt{a(a+b)}-a .
\end{array}
$$
Thus, $A F=a+x=\sqrt{a(a+b)}$
$$
=\sqrt{A D \cdot A B}=A C \text {. }
$$
Similarly, $B E=B C$.
Notice th... | 45^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 724,483 |
1. For the inscribed $\triangle ABC$ in a circle, $\angle A \neq 90^{\circ}, AB > AC$, draw the tangents to the circle at points $B$ and $C$ respectively, and let their intersection be $P$. Let $M$ be the midpoint of $BC$. Prove that $\angle BAM$ and $\angle PAC$ are equal or supplementary. | Let $P A$ intersect the circle $\Gamma$ at point $K$. Using $A B \cdot C K=A C \cdot B K$ and Ptolemy's theorem, we can get $\triangle A B M \backsim \triangle A K C$. When $\angle A=90^{\circ}$, $\angle B A M$ and $\angle P A C$ are supplementary. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,484 |
2. In Rt $\triangle A B C$, $C D$ is the altitude on the hypotenuse $A B$, $O_{1}$ and $O_{2}$ are the incenters of $\triangle A C D$ and $\triangle B C D$ respectively, the line $O_{1} O_{2}$ intersects $A C$ and $B C$ at points $E$ and $F$ respectively, and the rays $D O_{1}$ and $D O_{2}$ intersect $A C$ and $B C$ a... | Prompt: From Example 3, we get $C E = C D = C F$.
From the fact that points $C, M, D, N$ are concyclic, we get $C M = C N$.
Thus, $M N \parallel E F$.
Let $B C = a, C A = b, A B = c$.
Notice that $\frac{M N}{O_{1} O_{2}} = \frac{D M}{D O_{1}} = \frac{a + b + c}{a + b}$,
$\frac{M N}{E F} = \frac{C N}{C F} = \frac{C N}{C... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,485 |
3. In $\triangle A B C$, $A B>A C, \odot O$ is the incircle, $M, N, H$ are the points of tangency on sides $A B, A C, B C$ respectively, points $E, F$ are on sides $A B, A C$ respectively, and $E F \parallel B C, E F$ is tangent to $\odot O$ at point $K$. Let the line $M N$ intersect $E F$ at point $P, D$ is the midpoi... | Prompt: From Example 5, we know that $B F$, $C E$, and $K H$ are concurrent, and $K$, $O$, and $H$ are collinear. Observing the right trapezoid $P D H K$, it can be proven through calculation that: $P D = P K + D H$. Therefore, the circle with the right-angle leg as its diameter must be tangent to the slant leg, i.e., ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,486 |
4. $\odot O$ diameter $A B=2 R$. A moving point $P$ is on the perpendicular bisector of line segment $A O$, $P M$ and $P N$ are tangent to $\odot O$ at points $M$ and $N$ respectively. Prove: line $M N$ always passes through a certain fixed point. | Let the midpoint of $A O$ be $Q$, and $M N$ intersects $A B$ at point $G$. Let $G A=x$. Then $M, Q, O, N$ are concyclic.
By the power of a point theorem we get
$$
\begin{array}{l}
G A \cdot G B=G M \cdot G N=G Q \cdot G O \\
\Rightarrow x(x+2 R)=\left(x+\frac{1}{2} R\right)(x+R) \\
\Rightarrow x=R \Rightarrow G A=R .
\... | G A=R | Geometry | proof | Yes | Yes | cn_contest | false | 724,487 |
5. Given that $C D$ is the altitude on the hypotenuse $A B$ of the right triangle $\triangle A B C$, $\odot O$ is its circumcircle, $\odot O_{1}$ is internally tangent to arc $\overparen{A C}$ and tangent to $A B$ and $C D$, with $E$ being the tangency point on side $A B$; $\odot O_{2}$ is internally tangent to arc $\o... | Prompt: From Example 8, we get $A F=A C, B E=B C$. Then prove that $C E$ bisects $\angle A C D$, and $C F$ bisects $\angle D C B$. Thus, $\frac{A E}{E D} \cdot \frac{D F}{F B} \cdot \frac{B C}{C A}=\frac{A C}{C D} \cdot \frac{C D}{C B} \cdot \frac{B C}{C A}=1$. | 1 | Geometry | proof | Yes | Yes | cn_contest | false | 724,488 |
Second Question: Prove that for any integer $n(n \geqslant 4)$, there exists an $n$-degree polynomial
$$
f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}
$$
with the following properties:
(1) $a_{0}, a_{1}, \cdots, a_{n-1}$ are all positive integers;
(2) For any positive integer $m$ and any $k(k \geqslant 2)$ distinct ... | Proof 1 (i) When $n$ is even, let $n=2 t+2(t \geqslant 1)$.
Obviously, for any integer $x$, we have $41 x^{2}\left(x^{2}-1\right)$.
Indeed, when $x$ is even, $4 \mid x^{2}$; when $x$ is odd, $41\left(x^{2 t}-1\right)$.
Thus, for any integer $x$, we have $x^{n} \equiv x^{2}(\bmod 4)$.
Let $f(x)=x^{n}+4\left(x^{n-1}+x^{n... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,490 |
Third question: Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 4)$ be given positive real numbers, $a_{1}<a_{2}<\cdots<a_{n}$. For any positive real number $r$, the number of triples $(i, j, k)$ satisfying $\frac{a_{j}-a_{i}}{a_{k}-a_{j}}=r(1 \leqslant i<j<k \leqslant n)$ is denoted by $f_{n}(r)$. Prove: $f_{n}(r)<\frac{... | Proof 1 For any positive real number $r$, we have $\frac{a_{j}-a_{i}}{a_{k}-a_{j}}=r(1 \leqslant i<j<k \leqslant n)$,
which means $a_{k}=a_{j}+\frac{a_{j}-a_{i}}{r}$.
For a fixed $j \leqslant\left[\frac{n}{2}\right]$, by equation (1), each $i$ corresponds to at most one $k$, so the number of triples $(i, j, k)$ is
$$
s... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,491 |
Example 4 As shown in Figure 2, let $M$ be the centroid of $\triangle A B C$, and a line through $M$ intersects sides $A B$ and $A C$ at points $P$ and $Q$, respectively, and
$$
\frac{A P}{P B}=m, \frac{A Q}{Q C}=n \text {. }
$$
Then $\frac{1}{m}+\frac{1}{n}=$ | Solution 1 As shown in Figure 2, draw lines through points $C$ and $B$ parallel to $PQ$ intersecting $AD$ or its extension at points $E$ and $F$. Then,
$$
\begin{array}{l}
\frac{1}{m}=\frac{P B}{A P}=\frac{F M}{A M}, \\
\frac{1}{n}=\frac{Q C}{A Q}=\frac{E M}{A M} .
\end{array}
$$
Since $D$ is the midpoint of side $BC$... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,492 |
1. Given positive integers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a b+b c+c a+2(a+b+c)=8045, \\
a b c-a-b-c=-2 .
\end{array}\right.
$$
then $a+b+c=$ $\qquad$ | $$
-1.2012 .
$$
Note that
$$
\begin{array}{l}
(a+1)(b+1)(c+1) \\
=a b c+a b+b c+c a+a+b+c+1 \\
=8045+(-2)+1=8044 .
\end{array}
$$
Since $a, b, c$ are positive integers, we have
$$
a+1 \geqslant 2, b+1 \geqslant 2, c+1 \geqslant 2 \text{. }
$$
Therefore, 8044 can only be factored as $2 \times 2 \times 2011$. Hence $a... | 2012 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,493 |
2. In a class, there are two types of students: one type always lies, and the other type never lies. Each student knows what type the other students are. During a gathering today, each student has to state what type the other students are, and all students together said "liar" 240 times. At a similar gathering yesterda... | 2. 22 .
Consider four possible scenarios:
(1) If student $A$ is a liar and student $B$ is not a liar, then $A$ will say $B$ is a liar;
(2) If student $A$ is a liar and student $B$ is also a liar, then $A$ will say $B$ is not a liar;
(3) If student $A$ is not a liar and student $B$ is also not a liar, then $A$ will say... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 724,494 |
3. On the blackboard, it is written
$$
1!\times 2!\times \cdots \times 2011!\times 2012!\text {. }
$$
If one of the factorials is erased so that the remaining product equals the square of some positive integer, then the erased term is . $\qquad$ | 3. 1006 !.
Notice that the given product equals
$$
\begin{array}{l}
1^{2012} \times 2^{2011} \times \cdots \times 2011^{2} \times 2012^{1} \\
=\left(1^{1006} \times 2^{1005} \times 3^{1005} \times \cdots \times 2010^{1} \times \\
2011^{1}\right)^{2} \times 2^{1000} \times 1006!
\end{array}
$$ | 1006! | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,495 |
4. As shown in Figure 1, points $B$ and $E$ are on the sides $AD$ and $AC$ of $\triangle ACD$, respectively. $BC$ intersects $DE$ at point $F$. Given that $\triangle ABC$ and $\triangle AED$ are congruent, $AB = AE = 1$, and $AC = AD = 3$. Then the ratio of $S_{\text{quadrilateral } ABFE}$ to $S_{\triangle MDC}$ is $\q... | 4. $\frac{1}{6}$.
Notice that
$$
S_{\triangle A B F}=S_{\triangle A B F}=\frac{1}{2} h, S_{\triangle A D F}=\frac{3}{2} h .
$$
Therefore, $S_{\text {quadrilateral } A B F E}=h, S_{\triangle D C}=3 S_{\triangle A D E}=6 h$. Thus, the required ratio is $\frac{1}{6}$. | \frac{1}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,496 |
5. The positive integer $n$ has exactly 4 positive divisors (including 1 and $n$). It is known that $n+1$ is four times the sum of the other two divisors. Then $n=$ | 5.95.
Notice that a positive integer with exactly four positive divisors must be of the form $p^{3}$ or $p q$ (where $p$ and $q$ are primes, $p \neq q$).
In the first case, all positive divisors are $1, p, p^{2}, p^{3}$, then $1+p^{3}=4\left(p+p^{2}\right)$, but $p \nmid \left(1+p^{3}\right)$, which is a contradictio... | 95 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,497 |
7. As shown in Figure 2, the radii of circles $A$ and $B$ are both 1, and they are externally tangent to each other. The radii of circles $P, Q, R, S$ are all $r$, and circle $P$ is externally tangent to circles $A, B, Q, S$, circle $Q$ is externally tangent to circles $P, B, R$, circle $R$ is externally tangent to cir... | 7. $\frac{3+\sqrt{17}}{2}$
As shown in Figure 7,
let $M_{1} 、 M_{2}$, and
$M_{3}$ represent the
centers of circles $A 、 P 、 S$,
respectively. Suppose circle $A$
is tangent to circle $B$ at
point $T$. By the Pythagorean
theorem, we have
$$
\begin{array}{l}
\left|M_{1} M_{2}\right|^{2}-\left|M_{1} T\right|^{2} \\
=\left... | \frac{3+\sqrt{17}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,499 |
9. In a triangle with side lengths of $50, 120, 130$, take all points inside and outside the triangle that are at least 2 units away from at least one point on the triangle's sides. Then the area of the region formed by all the taken points is $\qquad$ (take $\left.\pi=\frac{22}{7}\right)$. | 9. $1182 \frac{4}{7}$.
The region is shown in Figure 8.
The initial triangle divides it into inner and outer parts. The part outside the triangle includes three sectors near the vertices, which can be combined to form a circle with a radius of 2; the remaining three rectangles are $2 \times 50, 2 \times 120, 2 \times... | 1182 \frac{4}{7} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,501 |
10. Given a positive integer $n$ that satisfies the following conditions:
(1) It is an eight-digit number, and all its digits are 0 or 1;
(2) Its first digit is 1;
(3) The sum of the digits in the even positions equals the sum of the digits in the odd positions.
How many such $n$ are there? | 10.35.
From the fact that the sum of the digits in the even positions equals the sum of the digits in the odd positions, we know that the number of 1s in the even positions equals the number of 1s in the odd positions.
Since the first digit is fixed, there are only three positions in the odd positions that can be fre... | 35 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,502 |
Example 5 As shown in Figure 4, if $P$ is any point inside the equilateral $\triangle A B C$, and perpendiculars are drawn from $P$ to the sides $A B$, $B C$, and $C A$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively, then $\frac{A D+B E+C F}{P D+P E+P F}=$ ( ).
(A) 2
(B) $\sqrt{3}$
(C) $2 \sq... | Solution 1 As shown in Figure 4, let the side length of the equilateral $\triangle ABC$ be $a$. Draw $A_{1} B_{1} \parallel AB, B_{2} C_{2} \parallel BC, C_{1} A_{2} \parallel CA$, intersecting sides $BC$, $CA$, and $AB$ at points $B_{1}$ and $C_{1}$, $A_{1}$ and $C_{2}$, $A_{2}$ and $B_{2}$, respectively. It is easy t... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,503 |
11. On an infinitely large chessboard, there is 1 chess piece on a small square, where each small square is $1 \mathrm{~cm} \times 1 \mathrm{~cm}$. The chess piece moves according to the following rules:
(1) On the 1st move, the chess piece moves one square north.
(2) On the $n$-th move, if $n$ is odd, the chess piece ... | 11.2.
All even-numbered moves are along the east-west direction, while all odd-numbered moves are along the north-south direction. The piece has traveled a total of $36 \mathrm{~cm}$ along the east-west direction and $42 \mathrm{~cm}$ along the north-south direction. The displacement along the east-west direction can ... | 11.2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,504 |
12. Given real numbers $a, b, c$ satisfy
$$
\frac{a(b-c)}{b(c-a)}=\frac{b(c-a)}{c(b-a)}=k>0,
$$
where $k$ is some constant. Then the greatest integer not greater than $k$ is . $\qquad$ | 12. 0 .
Let the substitution be $x=a b, y=b c, z=c a$. Then
$$
x-z=k(y-x), y-x=k(y-z) \text {. }
$$
Thus, $x-z=k^{2}(y-z)$. Therefore,
$$
\begin{array}{l}
(y-x)+(x-z)+(z-y)=0 \\
\Leftrightarrow(y-z)\left(k^{2}+k-1\right)=0(x \neq y \neq z) \\
\Rightarrow k^{2}+k-1=0 .
\end{array}
$$
Also, $k>0 \Rightarrow k=\frac{-1... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,505 |
13. Quadrilateral $A B C D$ has diagonals $A C$ and $B D$ intersecting at point $E$. If $A E=C E, \angle A B C=\angle A D C$, is quadrilateral $A B C D$ definitely a parallelogram? | II. 13. Yes.
As shown in Figure 9, construct quadrilateral ABCF.
Since $AC$ and $BF$ bisect each other, point $D$ lies on line $BE$.
If point $D$ is between points $E$ and $F$ (as shown in Figure 9), then
$$
\begin{array}{l}
\angle A D C=\angle A D E+\angle E D C \\
>\angle A F E+\angle E F C=\angle A F C=\angle A B C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,506 |
14. The roots of the equation $x^{2}-2 x-a^{2}-a=0$ are $\left(\alpha_{a}\right.$,
$$
\begin{array}{r}
\left.\beta_{a}\right)(a=1,2, \cdots, 2011) . \\
\text { Find } \sum_{a=1}^{2011}\left(\frac{1}{\alpha_{a}}+\frac{1}{\beta_{a}}\right) \text { . }
\end{array}
$$ | 14. The roots of the equation $x^{2}-2 x-a^{2}-a=0$ should satisfy
$$
\begin{array}{l}
\alpha_{a}+\beta_{a}=2, \alpha_{a} \beta_{a}=-\left(a^{2}+a\right) . \\
\text { Then } \sum_{a=1}^{2011}\left(\frac{1}{\alpha_{a}}+\frac{1}{\beta_{a}}\right)=-\sum_{a=1}^{2011} \frac{2}{a^{2}+a} \\
=-2 \sum_{a=1}^{2011}\left(\frac{1}... | -\frac{2011}{1006} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,507 |
15. Given that 15 rays share a common endpoint. Question: What is the maximum number of obtuse angles (considering the angle between any two rays to be the one not greater than $180^{\circ}$) that these 15 rays can form? | 15. First, it is explained that constructing 75 obtuse angles is achievable.
The position of the rays is represented by their inclination angles, with 15 rays placed near the positions of $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$. Within each group, the five rays are sufficiently close to each other (as shown in F... | 75 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,508 |
1. (40 points) Find all real roots of the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right)
$$ | 1. The original equation can be transformed into
$$
x^{4}+3 x^{3}+6 x^{2}+7 x+3=0 \text {. }
$$
Let $x=-1$. Then $1-3+6-7+3=0$.
Therefore, $x+1$ is a factor.
$$
\begin{array}{l}
\text { Hence } x^{4}+3 x^{3}+6 x^{2}+7 x+3 \\
=(x+1)\left(x^{3}+2 x^{2}+4 x+3\right) \\
=(x+1)^{2}\left(x^{2}+x+3\right)=0 .
\end{array}
$$
... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,509 |
Example 6 As shown in Figure 5, in $\triangle A B C$, $\angle B=\angle C$, point $D$ is on side $B C$, $\angle B A D=50^{\circ}$. Take point $E$ on $A C$ such that $\angle A D E=$ $\angle A E D$. Then the degree measure of $\angle E D C$ is ( ).
(A) $15^{\circ}$
(B) $25^{\circ}$
(C) $30^{\circ}$
(D) $50^{\circ}$ | Solution 1 Notice that
$$
\begin{array}{l}
\angle E D C=\angle A D C-\angle A D E \\
=(\angle B A D+\angle B)-\angle A E D \\
=\angle B A D+\angle B-(\angle E D C+\angle C) \\
=\angle B A D-\angle E D C .
\end{array}
$$
Therefore, $\angle E D C=\angle B A D \div 2=25^{\circ}$.
Solution 2 Let $\angle D A C=50^{\circ}$.... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,514 |
6. (40 points) Let $f(x)$ be a polynomial of degree 2010 such that $f(k)=-\frac{2}{k}$ for $k=1,2, \cdots, 2011$.
Find $f(2012)$. | 6. Let $g(x)=x f(x)+2$.
Since the degree of $f(x)$ is 2010, the degree of $g(x)$ is 2011.
$$
\text { Also, } g(k)=k f(k)+2=0 \text { for } k=1,2, \cdots, 2011, \text { then }
$$
$$
g(x)=\lambda(x-1)(x-2) \cdots(x-2011),
$$
where $\lambda$ is some real number.
Let $x=0$. Then $2=-2011! \cdot \lambda$.
Thus, $g(2012)=2... | -\frac{1}{503} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,515 |
7. (40 points) A cat caught 81 mice and arranged them in a circle, numbering them from $1 \sim 81$ in a clockwise direction. The cat starts counting from a certain mouse in a clockwise direction, continuously counting “$1, 2, 3$” and eating all the mice that are counted as 3. As the cat continues to count, the circle g... | 7. First, arrange the numbers of all the mice in 9 columns as follows.
\begin{tabular}{ccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\
19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 \\
28 & 29 & 30 & 31 & 32 & 33 & 34 & 35 & 36 \\
37 & 38 & 39 & 40 & 41 & 42 & 43 & 44 & 45 \\
4... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 724,516 |
8. (40 points) In $\triangle A B C$, it is given that $B C=A C$, $\angle B C A=90^{\circ}$, points $D$ and $E$ are on sides $A C$ and $A B$ respectively, such that $A D=A E$, and $2 C D=B E$. Let $P$ be the intersection of segment $B D$ and the angle bisector of $\angle C A B$. Find $\angle P C B$.
---
The translatio... | 8. Solution 1 As shown in Figure 14, let $F$ be the midpoint of side $BE$. Then $FC$ is parallel to $ED$.
Extend $ED$ to intersect the extension of $BC$ at point $Q$. Thus, $\triangle ABQ$ is an isosceles right triangle.
$$
\begin{aligned}
& \text{Then } \angle AQE \\
= & \angle ADE - 45^{\circ} \\
= & \angle AED - 45^... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,517 |
9. (40 points) From the 49 small squares of a $7 \times 7$ grid, select 21 to color, such that no four of the colored squares can form the four corners of any smaller grid.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
9. (40 points) From the 49 small squares of a $7 \times 7$ grid, select 21 to color, such that no four of the ... | 9. Figure 15 gives a coloring scheme.
As shown in Figure 16, construct 7 points and 7 lines. As long as the lines correspond to rows, and the points correspond to columns, a coloring scheme that meets the requirements of the problem is obtained. | not found | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,518 |
10. (40 points) Given positive integers $a$, $b$, and $c$ to 甲, 乙, and 丙 respectively, each person only knows their own number. They are told that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, and each is asked the following two questions:
(1) Do you know the value of $a+b+c$?
(2) Do you know the values of $a$, $b$, and $c$... | 10. First solve the equation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.
Assume $a \geqslant b \geqslant c$. If $c=3$, then $1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{3}{c}$.
Thus, $a=b=c=3$. Otherwise, $c=2$.
Then, $\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$.
Solving this, we get $a=b=4$ or $b=3, a=6$.
The sums co... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 724,519 |
1. Let the set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$. If the set $B=\{-1,3,5,8\}$ consists of the sums of all three-element subsets of $A$, then the set $A=$ $\qquad$ . | - 1. $\{-3,0,2,6\}$.
Obviously, in all three-element subsets of set $A$, each element appears 3 times. Therefore,
$$
\begin{array}{l}
3\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=(-1)+3+5+8=15 \\
\Rightarrow a_{1}+a_{2}+a_{3}+a_{4}=5 .
\end{array}
$$
Thus, the four elements of set $A$ are
$$
5-(-1)=6,5-3=2,5-5=0,5-8=-3 \tex... | \{-3,0,2,6\} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,520 |
3. Let $a, b$ be positive real numbers, and
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}, \\
(a-b)^{2}=4(a b)^{3} .
\end{array}
$$
Then $\log _{a} b=$ $\qquad$ | 3. -1 .
From $\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}$, we get $a+b \leqslant 2 \sqrt{2} a b$.
$$
\begin{array}{l}
\text { Also, }(a+b)^{2}=4 a b+(a-b)^{2} \\
=4 a b+4(a b)^{3} \\
\geqslant 4 \times 2 \sqrt{a b(a b)^{3}}=8(a b)^{2},
\end{array}
$$
which means $a+b \geqslant 2 \sqrt{2} a b$.
Thus, $a+b=2 \sqrt{2}... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,521 |
4. If $\cos ^{5} \theta-\sin ^{5} \theta<7\left(\sin ^{3} \theta-\cos ^{3} \theta\right)$ $(\theta \in[0,2 \pi))$, then the range of values for $\theta$ is $\qquad$ | 4. $\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)$.
The given inequality is equivalent to
$$
7 \sin ^{3} \theta+\sin ^{5} \theta>7 \cos ^{3} \theta+\cos ^{5} \theta \text {. }
$$
Since $f(x)=7 x^{3}+x^{5}$ is an increasing function on $(-\infty,+\infty)$, it follows that $\sin \theta>\cos \theta$.
$$
\text { Hence } 2 ... | \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,522 |
5. Now arrange for 7 students to participate in 5 sports events, requiring that students A and B cannot participate in the same event, each event must have participants, and each person can only participate in one event. The number of different arrangements that meet the above requirements is $\qquad$ (answer in number... | 5. 15000 .
According to the problem, there are two scenarios that meet the conditions:
(1) One project has 3 participants, with a total of
$$
C_{7}^{3} \times 5! - C_{5}^{1} \times 5! = 3600
$$
schemes;
(2) Two projects each have 2 participants, with a total of
$$
\frac{1}{2}\left(\mathrm{C}_{7}^{2} \mathrm{C}_{5}^{2... | 15000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,523 |
Example 7 As shown in Figure 6, in rectangle $ABCD$, it is known that the diagonal length is 2, and
$$
\begin{array}{l}
\angle 1=\angle 2= \\
\angle 3=\angle 4 .
\end{array}
$$
Then the perimeter of quadrilateral
$\mathrm{EFGH}$
is ( ).
(A) $2 \sqrt{2}$
(B) 4
(C) $4 \sqrt{2}$
(D) 6
(2010, Sichuan Province Junior High ... | Solution 1 According to the reflection relationship, as shown in Figure $6, \mathrm{Rt} \triangle I J K$'s hypotenuse is the perimeter of quadrilateral $E F G H$, and the right-angle sides are twice the length of the rectangle's sides, therefore, the perimeter of quadrilateral $E F G H$ is twice the length of the recta... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,524 |
6. In the tetrahedron $ABCD$, it is known that
$$
\begin{array}{l}
\angle ADB=\angle BDC=\angle CDA=60^{\circ}, \\
AD=BD=3, CD=2 .
\end{array}
$$
Then the radius of the circumscribed sphere of the tetrahedron $ABCD$ is | 6. $\sqrt{3}$.
As shown in Figure 3, let the circumcenter of tetrahedron $ABCD$ be $O$. Then point $O$ lies on the perpendicular line through the circumcenter $N$ of $\triangle ABD$ and perpendicular to the plane $ABD$.
By the given conditions, $N$ is the center of the equilateral $\triangle ABD$.
Let $P$ and $M$ be t... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,525 |
7. The line $x-2 y-1=0$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$, and $C$ is a point on the parabola such that $\angle A C B$ $=90^{\circ}$. Then the coordinates of point $C$ are $\qquad$ . | 7. $(1,-2)$ or $(9,-6)$.
Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right) 、 C\left(t^{2}, 2 t\right)$.
From $\left\{\begin{array}{l}x-2 y-1=0, \\ y^{2}=4 x,\end{array}\right.$ we get $y^{2}-8 y-4=0$.
Then $\left\{\begin{array}{l}y_{1}+y_{2}=8, \\ y_{1} y_{2}=-4 \text {. }\end{array}\right.$
Also, $x_{1}=2 ... | (1,-2) \text{ or } (9,-6) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,526 |
8. Given
$$
a_{n}=\mathrm{C}_{200}^{n}(\sqrt[3]{6})^{200-n}\left(\frac{1}{\sqrt{2}}\right)^{n}(n=1,2, \cdots, 95) \text {. }
$$
The number of integer terms in the sequence $\left\{a_{n}\right\}$ is $\qquad$ | 8. 15 .
Notice that $a_{n}=\mathrm{C}_{200}^{n} \times 3^{\frac{200-n}{3}} \times 2^{\frac{200-5 n}{6}}$.
To make $a_{n}(1 \leqslant n \leqslant 95)$ an integer, it must be that $\frac{200-n}{3}$ and $\frac{400-5 n}{6}$ are both integers, i.e., $61(n+4)$.
When $n=6 k+2(k=0,1, \cdots, 13)$, $\frac{200-n}{3}, \frac{400... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,527 |
9. (16 points) Let the function $f(x)=|\lg (x+1)|$, and real numbers $a, b (a<b)$ satisfy
$$
f(a)=f\left(-\frac{b+1}{b+2}\right), f(10 a+6 b+21)=4 \lg 2 \text {. }
$$
Find the values of $a, b$. | From the given, we have
$$
\begin{array}{l}
|\lg (a+1)|=\left|\lg \left(-\frac{b+1}{b+2}+1\right)\right| \\
=\left|\lg \frac{1}{b+2}\right|=|\lg (b+2)| .
\end{array}
$$
Thus, $a+1=b+2$ or $(a+1)(b+2)=1$.
Given $a1 \text {. }$
$$
Therefore, $f(10 a+6 b+21)$
$$
=\lg \left[6(b+2)+\frac{10}{b+2}\right]=4 \lg 2 \text {. }... | a=-\frac{2}{5}, b=-\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,528 |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=2 t-3(t \in \mathbf{R}, \text { and } t \neq \pm 1), \\
a_{n+1}=\frac{\left(2 t^{n+1}-3\right) a_{n}+2(t-1) t^{n}-1}{a_{n}+2 t^{n}-1}\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
(1) Find the general term formula of the ... | 10. (1) From the original equation, we have
$$
\begin{array}{l}
a_{n+1}=-\frac{2\left(t^{n+1}-1\right)\left(a_{n}+1\right)}{a_{n}+2 t^{n}-1}-1 . \\
\text { Then } \frac{a_{n+1}+1}{t^{n+1}-1}=\frac{2\left(a_{n}+1\right)}{a_{n}+2 t^{n}-1}=\frac{\frac{2\left(a_{n}+1\right)}{t^{n}-1}}{\frac{a_{n}+1}{t^{n}-1}+2} .
\end{arra... | a_{n}=\frac{2\left(t^{n}-1\right)}{n}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,529 |
11. (20 points) A line $l$ with a slope of $\frac{1}{3}$ intersects the ellipse $C$: $\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$ at points $A$ and $B$ (as shown in Figure 1), and $P(3 \sqrt{2}, \sqrt{2})$ is located above and to the left of line $l$.
(1) Prove that the incenter of $\triangle P A B$ lies on a fixed line;
(2) I... | 11. (1) Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
line $l: y=\frac{1}{3} x+m$.
Substituting equation (1) into the equation of ellipse $C$ and simplifying, we get
$$
2 x^{2}+6 m x+9 m^{2}-36=0.
$$
Then $x_{1}+x_{2}=-3 m, x_{1} x_{2}=\frac{9 m^{2}-36}{2}$,
$$
k_{P A}=\frac{y_{1}-\sqrt{2}}{x_{1}-3 \s... | \frac{117 \sqrt{3}}{49} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,530 |
One, (40 points) As shown in Figure 2, $P$ and $Q$ are the midpoints of the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$. If $\angle BPA = \angle DPA$, prove:
$$
\begin{array}{l}
\angle AQB \\
=\angle CQB .
\end{array}
$$ | As shown in Figure 4, extend line segments $DP$ and $AQ$ to intersect the circle at points $E$ and $F$ respectively. Then,
$$
\begin{array}{l}
\angle CPE \\
=\angle DPA \\
=\angle BPA.
\end{array}
$$
Since $P$ is the midpoint of line segment $AC$, we have
$$
\overparen{AB}=\overparen{CE}.
$$
Thus,
$$
\angle CDP=\angl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,531 |
II. (40 points) Prove: For any integer $n(n \geqslant 4)$, there exists an $n$-degree polynomial
$$
f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}
$$
with the following properties:
(1) $a_{0}, a_{1}, \cdots, a_{n-1}$ are all positive integers;
(2) For any positive integer $m$ and any $k(k \geqslant 2)$
distinct posi... | Let
$$
f(x)=(x+1)(x+2) \cdots(x+n)+2 \text {. }
$$
Expanding the right-hand side of (1) shows that $f(x)$ is an $n$-degree polynomial with positive integer coefficients and a leading coefficient of 1.
We now prove that $f(x)$ satisfies property (2).
For any integer $t$, since $n \geqslant 4$, among the $n$ consecutive... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,532 |
Three. (50 points) Let $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 4)$ be given positive real numbers, $a_{1}<a_{2}<\cdots<a_{n}$. For any positive real number $r$, the number of triples $(i, j, k)$ satisfying $\frac{a_{j}-a_{i}}{a_{k}-a_{j}}=r(1 \leqslant i<j<k \leqslant n)$ is denoted by $f_{n}(r)$. Prove:
$$
f_{n}(r)<\... | For the given $j(1j$, that is, $k$ has $n-j$ choices, so,
$$
g_{j}(r) \leqslant n-j \text {. }
$$
Thus, $g_{j}(r) \leqslant \min \{j-1, n-j\}$.
Therefore, when $n$ is even, let $n=2 m$. Then
$$
\begin{aligned}
& f_{n}(r)=\sum_{j=2}^{n-1} g_{j}(r)=\sum_{j=2}^{m-1} g_{j}(r)+\sum_{j=m}^{2 m-1} g_{j}(r) \\
\leqslant & \su... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,533 |
Four. (50 points) Let $A$ be a $3 \times 9$ grid, with each small cell filled with a positive integer. If the sum of all numbers in an $m \times n (1 \leqslant m \leqslant 3, 1 \leqslant n \leqslant 9)$ subgrid of $A$ is a multiple of 10, then it is called a "good rectangle"; if a $1 \times 1$ cell in $A$ is not contai... | First, we prove by contradiction that there are no more than 25 bad cells in $A$.
Assume the conclusion is not true. Then, in the grid $A$, there is at most 1 cell that is not a bad cell. By the symmetry of the grid, we can assume that all cells in the first row are bad cells.
Let the numbers filled in the $i$-th col... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,534 |
11. 1. Given that $d$ and $d^{\prime}\left(d^{\prime}>d\right)$ are two positive divisors of the positive integer $n$. Prove: $d^{\prime}>d+\frac{d^{2}}{n}$. | 11.1. Since $f=\frac{n}{d}, f^{\prime}=\frac{n}{d^{\prime}}$ are integers, and $f>f^{\prime}$, we have
$$
\begin{array}{l}
1 \leqslant f-f^{\prime}=\frac{n}{d}-\frac{n}{d^{\prime}}=\frac{\left(d^{\prime}-d\right) n}{d d^{\prime}} \\
<\frac{\left(d^{\prime}-d\right) n}{d^{2}} .
\end{array}
$$
This inequality is clearly... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,536 |
11.2. On the side $BC$ of quadrilateral $ABCD$ ($\left.\angle A<90^{\circ}\right)$, take a point $T$ such that $\triangle A T D$ is an acute triangle. Let $O_{1}$, $O_{2}$, and $O_{3}$ be the circumcenters of $\triangle A B T$, $\triangle D A T$, and $\triangle C D T$, respectively. Prove that the orthocenter of $\tria... | 11.2. From $O_{1} O_{2}$ and $O_{3} O_{2}$ being the perpendicular bisectors of segments $A T$ and $D T$, respectively, we know that
$$
\angle A O_{1} O_{2}=\angle T O_{1} O_{2}=\angle T B A
$$
(because $\angle T O_{1} A$ is the central angle corresponding to the circumcircle of $\triangle A B T$).
Similarly, from
$$
\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,537 |
11.3. In the Academy, 999 academicians discuss several scientific issues. For each issue, exactly three academicians are interested. Every two academicians have exactly one issue they are both interested in. Prove: There exist 250 scientific issues such that each academician is interested in at most one of them. | 11.3. If an academician is interested in two questions, they are called "intersecting".
Let the discussed questions have at most $k$ pairwise non-intersecting questions: $T_{1}, T_{2}, \cdots, T_{k}$. Let $k \leqslant 249, S$ be the set of academicians not interested in these $k$ questions. Then
$$
s=|S|=999-3 k \geqs... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,538 |
11.4.10 Ten cars travel along a road that passes through several towns. It is known that each car travels through the towns at the same speed, and travels at a different uniform speed outside the towns (these speeds can be different for different cars). There are 2011 flags set up along the road. It is known that there... | 11.4. Introduce the spatial rectangular coordinate system Oxyt.
Let the initial position of the first car at time zero be point $M$, and each point $A$ on the road corresponds to a point $T_{A}\left(x_{A}, y_{A}\right)$ in the plane $x O y$, where $x_{A}$ is the sum of distances within towns along the road segment $A ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,539 |
11.5. Given that $F(x)$ and $G(x)$ are two real-coefficient cubic polynomials with leading coefficients of 1, the equations
$$
F(x)=0, G(x)=0, F(x)=G(x)
$$
have eight distinct real roots. Prove: The largest and smallest of these eight roots cannot both be roots of $F(x)=0$. | 11.5. Note that $F(x)$ and $G(x)$ are cubic polynomials, and $F(x)-G(x)$ is at most quadratic. Therefore, they have at most eight roots in total. Thus, each root is a simple root and $F(x)-G(x)$ is quadratic.
Let $a, b$ be the smallest and largest roots, respectively, and satisfy
$$
F(a)=F(b)=0 \text {. }
$$
Note that... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,540 |
11.6. There are more than $n^{2}$ stones on the table. Betja and Watt play a game: Betja starts, and the two take turns to remove stones from the table, each time taking a number of stones that is either a prime number less than $n$, a positive integer multiple of $n$, or 1. The rule is: the one who takes the last ston... | 11.6. Assume Watt has a winning strategy.
Let there be $d$ stones on the table initially, and $d$ divided by $n$ leaves a remainder $r$. Then
$1 \leqslant r \leqslant n-1$.
Betya first takes any multiple of $n$ stones, leaving $a_{k}=r+n k(0 \leqslant k \leqslant n-1)$ stones on the table.
Suppose Watt, under the win... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,541 |
11.7. For a positive integer $a$, $P(a)$ denotes the largest prime factor of $a^{2}+1$. Prove: there exist infinitely many different triples of distinct positive integers $(a, b, c)$ such that $P(a)=P(b)=P(c)$. | 11.7. First, we prove a lemma.
Lemma: Let $p>2$ be a prime number, and let $a<p$ be a positive integer such that $p \mid (a^{2}+1)$.
Then $P(a)=P(p-a)=p$.
Proof: Note that
$$
(p-a)^{2}+1 \equiv a^{2}+1 \equiv 0(\bmod p),
$$
and they are both less than $p^{2}$.
Therefore, $P(a)=P(p-a)=p$.
Returning to the original pro... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,542 |
11.8. Given a non-isosceles $\triangle ABC$, $N$ is the midpoint of the arc $\overparen{BAC}$ of its circumcircle, $M$ is the midpoint of side $BC$, and $I_{1}, I_{2}$ are the incenters of $\triangle ABM$ and $\triangle ACM$, respectively. Prove that $I_{1}, I_{2}, A, N$ are concyclic. | 11.8. Let $I_{2}^{\prime}$ be the reflection of point $I_{2}$ over $M N$.
$$
\begin{array}{l}
\text { Then } \angle B M I_{1}+\angle B M I_{2}^{\prime} \\
=\angle B M I_{1}+\angle C M I_{2} \\
=\frac{1}{2}(\angle B M A+\angle C M A)=90^{\circ} \\
=\angle B M N .
\end{array}
$$
Hence, $M I_{1}$ and $M I_{2}^{\prime}$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,543 |
1. Let $\odot O, \odot I$ be the circumcircle and incircle of Rt $\triangle ABC$, $R, r$ be the radii of these circles, and $J$ be the point symmetric to the right-angle vertex $C$ with respect to $I$. Find the length of $OJ$. | 1. Obviously, the circle $\odot J$ with radius $2r$ is tangent to $AC$ and $BC$.
Below, we prove that $\odot J$ is tangent to $\odot O$.
As shown in Figure 1, consider the circle $\omega$ that is tangent to $AC$ and $BC$ at points $P$ and $Q$, respectively, and is internally tangent to $\odot O$ at point $T$.
Next, we ... | R - 2r | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,544 |
2. In two equal circles $\omega_{1}$ and $\omega_{2}$, circle $\omega_{1}$ passes through the center $O$ of circle $\omega_{2}$. $\triangle ABC$ is inscribed in circle $\omega_{1}$, and lines $AC$ and $BC$ are tangent to circle $\omega_{2}$. Prove:
$$
\cos A + \cos B = 1 \text{. }
$$ | 2. As shown in Figure 2, let $P$ be the antipode of point $O$ on circle $\omega_{1}$, and $A^{\prime}$ be the point of tangency of $A C$ with circle $\omega_{2}$.
Since $C O$ is the angle bisector of $\angle A C B$, points $A$ and $B$ are symmetric with respect to line $O P$.
$$
\begin{array}{l}
\text { Then } \frac{|... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,545 |
3. Two convex polygons $A_{1} A_{2} \cdots A_{n}$ and $B_{1} B_{2} \cdots B_{n}$ $(n \geqslant 4)$, any side of the first polygon is greater than the corresponding side of the second polygon. Question: Can it happen that any diagonal of the second polygon is greater than the corresponding diagonal of the first polygon? | 3. First, prove a lemma.
Lemma: Let $\triangle A B C$ and $\triangle A B C^{\prime}$ satisfy $A C > A C^{\prime}, B C > B C^{\prime}$. Then for any point $K$ on the line segment $A B$, $C K > C^{\prime} K$.
Proof: From the conditions, points $A, B, C^{\prime}$ are on the same side of the perpendicular bisector of lin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,547 |
4. Two points projected onto the sides of a quadrilateral fall on two concentric circles of different sizes (the projection of each point forms a cyclic quadrilateral). Prove: The quadrilateral is a parallelogram. | 4. Let point $P$ have its projection on a side lying on the circumference of $\odot O$, and point $P^{\prime}$ be symmetric to point $P$ with respect to $O$. In this case, the projection of point $P^{\prime}$ on the side also lies on the same circumference, and $P$ and $P^{\prime}$ are the foci of a certain conic secti... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,548 |
5. In Rt $\triangle A B C$, $\angle B=90^{\circ}$, $B H$ is the altitude, the incircle of $\triangle A B H$ $\odot I_{1}$ touches sides $A B$ and $A H$ at points $H_{1}$ and $B_{1}$, respectively, and the incircle of $\triangle C B H$ $\odot I_{2}$ touches sides $C B$ and $C H$ at points $H_{2}$ and $B_{2}$, respective... | 5. As shown in Figure 4, let $\angle A C B=\alpha$.
$$
\text { Then } \frac{I_{1} H_{1}}{I_{2} H_{2}}=\frac{A B}{B C}=\tan \alpha \text {. }
$$
Since $I_{1} H_{1} \perp A B, I_{2} H_{2} \perp B C$, the projections of these segments on $A C$ are equal to $I_{1} H_{1} \cos \alpha$ and $I_{2} H_{2} \sin \alpha$, respecti... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,549 |
6. Given that the incircle $\odot I$ of $\triangle A B C$ touches its sides at points $A^{\prime}, B^{\prime}, C^{\prime}$, and the orthocenters of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ coincide. Is $\triangle A B C$ an equilateral triangle? | 6. As shown in Figure 5, assume $\triangle A B C$ is not an equilateral triangle.
Let $O$ be the circumcenter of $\triangle A B C$, and $H$ be the orthocenter of both $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$. Let $A^{\prime \prime}, B^{\prime \prime}, C^{\prime \prime}$ be the second intersec... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,550 |
7. Two regular polyhedra $P$ and $Q$ are each divided into two parts by a plane. One part of $P$ and one part of $Q$ are then glued together along the cutting planes. Can a regular polyhedron be obtained that is different from either of the original polyhedra? If so, how many faces does this regular polyhedron have? | 7. Let $R$ be the resulting polyhedron. Clearly, part of the polyhedron $P$ includes a vertex $A$ (which is not on the cutting plane), and the polyhedral angle of $P$ at this vertex is also the polyhedral angle of $R$ at the same vertex, meaning that polyhedra $P$ and $R$ are similar.
Similarly, $Q$ is also similar to ... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,551 |
8. Circle $\Gamma$ is the circumcircle of $\triangle A B C$. On the sides of $\triangle A B C$, three points $A_{1} 、 B_{1} 、 C_{1}$ are given, and then the original triangle is erased. Prove: $\triangle A B C$ can be restored if and only if the lines $A A_{1} 、 B B_{1} 、 C C_{1}$ intersect at one point. | 8. First, prove a lemma.
Lemma For the triangles $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ inscribed in the circle $\Gamma$, their corresponding sides intersect at points $A_{1}, B_{1}, C_{1}$. Then
$$
\left(\frac{A C_{1}}{C_{1} B} \cdot \frac{B A_{1}}{A_{1} C} \cdot \frac{C B_{1}}{B_{1} A}\ri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,552 |
1. Given a sequence of numbers
$$
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \cdots, \frac{k}{1}, \frac{k-1}{2}, \cdots, \frac{1}{k} \text {. }
$$
In this sequence, the index of the 40th term that equals 1 is ( ).
(A) 3120
(B) 3121
(C) 3200
(D) 3201 | - 1. B.
For the terms where the sum of the numerator and denominator is $k+1$, we denote them as the $k$-th group. According to the arrangement rule, the 40th term with a value of 1 should be the 40th number in the $2 \times 40-1=79$ group, with the sequence number being
$$
\begin{array}{l}
(1+2+\cdots+78)+40 \\
=\fra... | 3121 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,553 |
2. In $\triangle A B C$, it is known that $A B=a^{2}, B C=4 a$, $A C=b^{2}-4$, and $a, b$ are odd numbers greater than 3. Then the relationship between $a$ and $b$ is ( ).
(A) $a>b$
(B) $a<b$
(C) $a=b$
(D) cannot be determined | 2. C.
It is known that $a^{2}>4 a$. By the triangle inequality, we have
$$
\begin{array}{l}
a^{2}-4 a<b^{2}-4<a^{2}+4 a \\
\Rightarrow(a-2)^{2}<b^{2}<(a+2)^{2} \\
\Rightarrow a-2<b<a+2 .
\end{array}
$$
Since $a$ and $b$ are positive odd numbers, we have $b=a$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,554 |
3. Given that $a$ and $b$ are integers, $c$ is a prime number, and
$$
(a+b)^{4}=c-|a-b| \text {. }
$$
Then the number of ordered pairs $(a, b)$ is $(\quad)$.
(A) 2
(B) 4
(C) 6
(D) 8 | 3. C.
From the given, we have $(a+b)^{4}+|a-b|=c$.
Since $a+b$ and $a-b$ have the same parity, the left side of the equation is even. Therefore, $c$ is the even prime number 2.
$$
\begin{array}{l}
\text { Hence }(a+b, a-b) \\
=(0,-2),(0,2),(1,-1),(1,1), \\
\quad(-1,-1),(-1,1) .
\end{array}
$$
Therefore, the ordered p... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,555 |
4. In $\triangle A B C$, it is known that $A D$ is the angle bisector of $\angle B A C$, and $A D=2$. Then $A B+A C(\quad$.
(A) greater than 4
(B) less than 4
(C) not greater than 4
(D) not less than 4 | 4. A.
As shown in Figure 1, since $\angle A D C > \angle B$, we can construct $\angle A D E$ $=\angle B$ within $\angle A D C$, where $D E$ intersects $A C$ at point $E$. It is easy to see that
$$
\begin{array}{l}
\triangle A B D \backsim \triangle A D E \\
\Rightarrow \frac{A B}{A D}=\frac{A D}{A E} \\
\Rightarrow A... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,556 |
5. The equation $\left(x^{2}-4\right)^{2}=x+4$ has ( ) real solutions.
(A) 1
(B) 2
(C) 3
(D) 4 | 5. D.
Let $x^{2}-4=y$.
Then $x+4=y^{2}$.
$$
\begin{array}{l}
\text { (1) }+ \text { (2) gives } \\
x^{2}+x=y^{2}+y \\
\Rightarrow(x-y)(x+y+1)=0 \\
\Rightarrow y=x \text { or } y=-x-1 .
\end{array}
$$
When $y=x$, substituting into equation (1) yields $x=\frac{1 \pm \sqrt{17}}{2}$;
When $y=-x-1$, substituting into equa... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,558 |
6. In the Cartesian coordinate system, with the origin $O$ as the center, draw $\odot O$. Let $\odot O$ intersect the positive half-axis of the $x$-axis at point $P$, and $D(6,8)$ lies on $\odot O$. Points $E$ and $F$ lie on the line segment $OP$ (not coinciding with points $O$ and $P$). Connect $DE$ and $DF$ and exten... | 6. A.
As shown in Figure 2, draw $DM \perp EF$ at point $M$, extend $DM$ to intersect $\odot O$ at point $N$, and connect $ON$ to intersect $BC$ at point $Q$.
Then $DM$ bisects $\angle EDF$.
Thus, $\overparen{CN}=\overparen{BN} \Rightarrow ON \perp BC$.
It is easy to see that points $G, M, Q, N$ are concyclic.
Theref... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,559 |
1. Given $x=\frac{5}{\sqrt{7}+\sqrt{2}}$. Then $\sqrt{2}$ can be expressed as a rational-coefficient cubic polynomial in terms of $x$ as $\qquad$ . | 2. $-\frac{1}{10} x^{3}+\frac{13}{10} x$.
Notice that
$$
x=\frac{5}{\sqrt{7}+\sqrt{2}}=\sqrt{7}-\sqrt{2} \Rightarrow x+\sqrt{2}=\sqrt{7} \text {. }
$$
Square both sides and simplify to get $x^{2}=5-2 \sqrt{2} x$.
Multiply both sides of the above equation by $x$ to get
$$
\begin{array}{l}
x^{3}=5 x-2 \sqrt{2} x^{2}=5 x... | \sqrt{2}=-\frac{1}{10} x^{3}+\frac{13}{10} x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,560 |
2. Function
$$
y=\left|\sqrt{x^{2}+4 x+5}-\sqrt{x^{2}+2 x+5}\right|
$$
The maximum value is $\qquad$. | 2. $\sqrt{2}$.
Notice that
$$
\begin{array}{l}
y= \mid \sqrt{(x+2)^{2}+(0-1)^{2}}- \\
\sqrt{(x+1)^{2}+(0-2)^{2}} \mid .
\end{array}
$$
The above expression represents the absolute value of the difference in distances from point $P(x, 0)$ to points $A(-2,1)$ and $B(-1,2)$ (as shown in Figure 3). Therefore,
$$
\begin{... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,561 |
$$
\begin{array}{l}
\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{199}-\frac{1}{200}}{\frac{1}{201^{2}-1^{2}}+\frac{1}{202^{2}-2^{2}}+\cdots+\frac{1}{300^{2}-100^{2}}} \\
= \\
\end{array}
$$ | 3. 400 .
Original expression
$$
\begin{array}{l}
=\frac{\left(1+\frac{1}{2}+\cdots+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{200}\right)}{\frac{1}{202 \times 200}+\frac{1}{204 \times 200}+\cdots+\frac{1}{400 \times 200}} \\
=\frac{\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}}{\frac{1}{400... | 400 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,562 |
4. Given Rt $\triangle A B C$ and Rt $\triangle A D C$ share the hypotenuse $A C, M$ and $N$ are the midpoints of sides $A C$ and $B D$ respectively. If $\angle B A C=30^{\circ}, \angle C A D=45^{\circ}, M N=1$, then the length of $A C$ is | 4. $2(\sqrt{6} \pm \sqrt{2})$.
In two scenarios:
(1) When points $B$ and $D$ are on the same side of line $A C$, as shown in Figure 4, connect $D M$ and $B M$.
Then $M D = M B = \frac{1}{2} A C$, and $D M \perp A C$.
Thus, $\triangle B C M$ is an equilateral triangle.
Therefore, $\angle D M B = \angle D M C - \angle B... | 2(\sqrt{6} \pm \sqrt{2}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,563 |
One, (20 points) Given that $m$ is a real number, the function
$$
f(x)=m x^{2}+4 x-5-m
$$
has a root in the interval $[-2,2]$. Find the range of real values for $m$. | When $m=0$, the function has one root $\frac{5}{4}$.
When $m \neq 0$, the discriminant
$$
\Delta=16+4 m(m+5)=4(m+1)(m+4) \text{. }
$$
(1) When $\Delta=0$, $m=-1$ or -4.
If $m=-1$, the function has one root which is 2.
If $m=-4$, the function has one root which is $\frac{1}{2}$.
(2) When $\Delta>0$ and the function has... | m \geqslant -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,564 |
In $\triangle A B C$, it is known that $\angle C=$ $90^{\circ}, B C=2 A C, C D \perp A B$ at point $D, E$ is the midpoint of $B C$, connect $A E$ and $C D$ intersect at point $F$, point $G$ is on $A B$, and $B G=C F$. Prove: $E F=2 E G$.
---
The translation maintains the original text's formatting and structure. | $$
\begin{array}{l}
\text{As } D C \perp A B, \text{ so, } \angle A C F=\angle B. \\
\text{Also, } A C=\frac{1}{2} B C=B E, C F=B G, \text{ then} \\
\triangle A C F \cong \triangle E B G \\
\Rightarrow \angle B G E=\angle C F A=\angle E F D \\
\Rightarrow D, F, E, G \text{ are concyclic} \\
\Rightarrow \angle G E F=\an... | E F=2 E G | Geometry | proof | Yes | Yes | cn_contest | false | 724,565 |
Three. (25 points) If a rational number $m$ can be expressed in the form $3 x^{2}-8 x y+6 y^{2} (x, y$ are rational numbers $)$, then $m$ is called a "good number". Question: Are the product and quotient of two good numbers also good numbers? Why? | Three, All are good numbers.
Reason: Notice
$$
\begin{aligned}
m & =3 x^{2}-8 x y+6 y^{2} \\
& =\left(x^{2}-4 x y+4 y^{2}\right)+\left(2 x^{2}-4 x y+2 y^{2}\right) \\
& =(x-2 y)^{2}+2(x-y)^{2} .
\end{aligned}
$$
Let \( x-2 y=a, x-y=b \). Then \( m=a^{2}+2 b^{2} \) is a good number.
For any two good numbers \( c \) and... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,566 |
1. A real-coefficient polynomial $P(x)$ of degree not exceeding 2011 takes integer values for any integer $x$, and the remainders when $P(x)$ is divided by $x-1, x-2, \cdots, x-2011$ are $1, 2, \cdots, 2011$ respectively. Then $\max _{x \in \{-1, -2, \cdots, -2011\}}|P(x)|$ has the minimum value of $\qquad$ | $-1.2011$
First, $P(x)-x$. can be divided by $x-1, x-2, \cdots$, $x-2011$ respectively, so we can assume
$$
P(x)-x=q(x-1)(x-2) \cdots(x-2011) \text {, }
$$
where, $q \in \mathbf{Q}$.
If $q \neq 0$, then
$$
\begin{array}{l}
|P(-2011)|=\left|-2011-q \cdot \frac{4022!}{2011!}\right| \\
\geqslant\left|q \cdot \frac{4022!}... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,567 |
Example 1 As shown in Figure 1, in $\triangle ABC$, $\angle BAC = \angle BCA = 44^{\circ}$, and $M$ is a point inside $\triangle ABC$ such that $\angle MCA = 30^{\circ}$ and $\angle MAC = 16^{\circ}$. Find the measure of $\angle BMC$.
(2005, Beijing Middle School Mathematics Competition (Grade 8)) | As shown in Figure 1, draw $B D \perp A C$, with the foot of the perpendicular at $D$. Extend $C M$ to intersect $B D$ at point $O$, and connect $O A$.
It is easy to see that $B D$ is the axis of symmetry of $\triangle A B C$.
$$
\begin{array}{l}
\text { Then } \angle O A C=\angle M C A=30^{\circ} \\
\Rightarrow \angle... | 150^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,568 |
Example 2 As shown in Figure 2, in $\triangle A B C$, $\angle A B C$ $=46^{\circ}, D$ is a point on side $B C$, $D C=A B$,
$\angle D A B=21^{\circ}$. Try to find
the degree measure of $\angle C A D$. | Solve: Fold $\triangle A B D$ along the line $A D$ to get $\triangle A E D$. Then
$$
\begin{array}{l}
\angle E A D=21^{\circ}, A E=A B, D E=B D . \\
\text { It is easy to see that, } \angle A D C=21^{\circ}+46^{\circ}=67^{\circ} \text {. } \\
\text { Therefore, } \angle A D E=\angle A D B \\
=180^{\circ}-67^{\circ}=113... | 67^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,569 |
4. In rectangle $A B C D$, $A B=12, A D=3, E, F$ are points on $A B, D C$ respectively. Then the minimum length of the broken line $A F E C$ is $\qquad$
(2009, National Junior High School Mathematics League Sichuan Preliminary) | As shown in Figure 12, construct the symmetric points $A_{1}$ and $C_{1}$ of points $A$ and $C$ with respect to $DC$ and $AB$ respectively. Connect $A_{1}C_{1}$, which intersects $AB$ and $DC$ at points $E_{1}$ and $F_{1}$ respectively. It is easy to know that when points $E$ and $F$ coincide with $E_{1}$ and $F_{1}$ r... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,570 |
1. Given that $x$ is a positive integer, and $2011-x$ is a perfect cube. Then the minimum value of $x$ is $\qquad$ . | 1. 283.
From $12^{3}=1728<2011<13^{3}=2197$, we know the minimum value of $x$ is
$$
2011-1728=283 \text {. }
$$ | 283 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,571 |
2. Given $a<b<0$, and $\frac{a}{b}+\frac{b}{a}=6$. Then $\left(\frac{a+b}{a-b}\right)^{3}=$ $\qquad$ | 2. $2 \sqrt{2}$.
$$
\begin{array}{l}
\text { Given } \frac{a}{b}+\frac{b}{a}=6 \Rightarrow a^{2}+b^{2}=6 a b \\
\Rightarrow(a+b)^{2}=8 a b,(a-b)^{2}=4 a b .
\end{array}
$$
Since $a<b<0$, we have,
$$
\begin{array}{l}
a+b=-2 \sqrt{2 a b}, \\
a-b=-2 \sqrt{a b} . \\
\text { Therefore }\left(\frac{a+b}{a-b}\right)^{3}=(\sq... | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,572 |
3. As shown in Figure $2, \triangle A B C$'s exterior angle $\angle A C D$'s angle bisector $C P$ intersects with the angle bisector $B P$ of the interior angle $\angle A B C$ at point $P$. If $\angle B P C=\alpha$, then $\angle C A P=$ $\qquad$ (express in terms of $\alpha$).
Figure 2 | $$
\text { 3. } 90^{\circ}-\alpha \text {. }
$$
It is easy to know that $P$ is the excenter of $\triangle A B C$. Then
$$
\begin{array}{l}
\angle B P C=\angle P C D-\angle P B C=\frac{1}{2} \angle B A \\
\Rightarrow \angle B A C=2 \alpha \\
\Rightarrow \angle C A P=\frac{180^{\circ}-2 \alpha}{2}=90^{\circ}-\alpha .
\e... | 90^{\circ}-\alpha | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,573 |
4. As shown in Figure 3, in $\triangle A B C$, let $E$, $F$, and $P$ be on sides $B C$, $C A$, and $A B$ respectively. Given that $A E$, $B F$, and $C P$ intersect at a point $D$, and $\frac{A D}{D E}+\frac{B D}{D F}+\frac{C D}{D P}=n$. Then $\frac{A D}{D E} \cdot \frac{B D}{D F} \cdot \frac{C D}{D P}=$ | \begin{array}{l}\text { 4. } n+2 \text {. } \\ \text { Let } S_{\triangle B D C}=x, S_{\triangle C D A}=y, S_{\triangle A D B}=z \text {. Then } \\ \frac{A D}{D E}=\frac{z+y}{x}, \frac{B D}{D F}=\frac{x+z}{y}, \frac{C D}{D P}=\frac{x+y}{z} . \\ \text { Therefore, } \frac{A D}{D E} \cdot \frac{B D}{D F} \cdot \frac{C D}... | n+2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,574 |
One. (20 points) Divide a cube with an edge length of a positive integer into 99 smaller cubes, among which, 98 smaller cubes are unit cubes. Find the surface area of the original cube. | Let the side length of the original cube be $x$, and the side lengths of the 99 smaller cubes be 1 and $y(x>y>1)$. Then
$$
\begin{array}{l}
x^{3}-y^{3}=98 \\
\Rightarrow(x-y)\left[(x-y)^{2}+3 x y\right]=98 \\
\Rightarrow(x-y) \mid 98=7^{2} \times 2 \\
\Rightarrow x-y=1,2,7,14,49,98 . \\
\text { By }(x-y)^{3}<x^{3}-y^{3... | 150 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,575 |
II. (25 points) As shown in Figure 4, given that $AB$ is the diameter of $\odot O$, chord $CD$ intersects $AB$ at point $E$. A tangent to $\odot O$ is drawn from point $A$ and intersects the extension of $CD$ at point $F$. Given $AC=8$, $CE:ED=6:5$, and $AE:EB=2:3$. Find the length of $AB$ and the value of $\tan \angle... | Let $C E=6 x, E D=5 x, A E=2 y$, $E B=3 y, D F=z$.
By the intersecting chords theorem, we have
$$
A E \cdot B E=C E \cdot E D \Rightarrow y=\sqrt{5} x \text {. }
$$
By the secant-tangent theorem, we have
$$
A F^{2}=D F \cdot C F=z(z+11 x) \text {. }
$$
By the Pythagorean theorem, we have
$$
\begin{array}{l}
A F^{2}=E... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,576 |
Three. (25 points) As shown in Figure 5, any line passing through point $F(0,1)$ intersects the parabola $y=\frac{1}{4} x^{2}$ at two points
$$
\begin{array}{l}
M\left(x_{1}, y_{1}\right), \\
N\left(x_{2}, y_{2}\right) \\
\left(x_{1}0\right).
\end{array}
$$
Figure 5
Let the projections of points $M, N, F$ on the line ... | (1) It is easy to know, $F_{1} M_{1} \cdot F_{1} N_{1}=-x_{1} x_{2}=4$.
And $F F_{1}=2$, then
$$
\begin{array}{l}
F_{1} M_{1} \cdot F_{1} N_{1}=F_{1} F^{2} \\
\Rightarrow \operatorname{Rt} \triangle M_{1} F F_{1} \backsim \operatorname{Rt} \triangle N_{1} F F_{1} \\
\Rightarrow \angle M_{1} F N_{1}=\angle M_{1} F F_{1}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,577 |
1. If $f(x)=x^{3}-3 x+m$ takes any three numbers $a, b, c$ in the interval $[0,2]$, there exists a triangle with side lengths $f(a), f(b), f(c)$, then the range of $m$ is $\qquad$ | 1. $(6,+\infty)$.
Notice that, $f^{\prime}(x)=3(x+1)(x-1)$.
When $x \in[0,1)$, $f^{\prime}(x)0, f(x)$ is increasing on (1,2].
Thus, when $x \in[0,2]$,
$$
\begin{array}{l}
f(x)_{\min }=f(1)=m-2>0, \\
f(2)=2+m>f(0)=m .
\end{array}
$$
By the given condition $2 f(1)>f(2)$.
Therefore, $m>6$. | (6,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,578 |
In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and $G$ is the centroid of $\triangle A B C$. If
$$
a \overrightarrow{G A}+b \overrightarrow{G B}+\frac{\sqrt{3}}{3} c \overrightarrow{G C}=0 \text {, }
$$
then $\angle A=$ . $\qquad$ | 2. $30^{\circ}$.
Notice that,
$$
\begin{array}{l}
\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=\mathbf{0} \\
\overrightarrow{G A}=-\overrightarrow{G B}-\overrightarrow{G C} .
\end{array}
$$
Then $a(-\overrightarrow{G B}-\overrightarrow{G C})+b \overrightarrow{G B}+\frac{\sqrt{3}}{3} c \overrightarro... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,579 |
3. Given positive real numbers $a(a \neq 1) 、 x 、 y$ satisfy
$$
\log _{a}^{2} x+\log _{a}^{2} y-\log _{a}(x y)^{2} \leqslant 2,
$$
and $\log _{a} y \geqslant 1$. Then the range of $\log _{a} x^{2} y$ is | 3. $[-1,2 \sqrt{5}+3]$.
Let $u=\log _{a} x, v=\log _{a} y$. Then $z=\log _{a} x^{2} y=2 \log _{a} x+\log _{a} y=2 u+v$. From the given conditions, we have
$$
\left\{\begin{array}{l}
(u-1)^{2}+(v-1)^{2} \leqslant 4, \\
v \geqslant 1 .
\end{array}\right.
$$
Then the point $(u, v)$ corresponds to the shaded region in Fi... | [-1,2 \sqrt{5}+3] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,580 |
5. In the Cartesian coordinate system, let point $A(-8,3)$, $B(-4,5)$, and moving points $C(0, n)$, $D(m, 0)$. When the perimeter of quadrilateral $A B C D$ is minimized, the ratio $\frac{m}{n}$ is ( ).
(A) $-\frac{2}{3}$
(B) -2
(C) $-\frac{3}{2}$
(D) -3 | As shown in Figure 13, let the symmetric point of point $A$ with respect to the $x$-axis be $A^{\prime}$, and the symmetric point of point $B$ with respect to the $y$-axis be $B^{\prime}$. Then $A(-8,-3)$, $B^{\prime}(4,5)$.
Therefore, when points $C$ and $D$ are both on the line $A^{\prime} B^{\prime}$, the perimeter... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,581 |
4. Let the function $y=f(x)$ be defined on $\mathbf{R}$, and have an inverse function $f^{-1}(x)$. It is known that the inverse function of $y=f(x+1)-2$ is $y=f^{-1}(2 x+1)$, and $f(1)=4$. If $n \in \mathbf{N}_{+}$, then $f(n)=$ $\qquad$ | $$
\begin{array}{l}
\text { 4. } 3+\left(\frac{1}{2}\right)^{n-1} \text {. } \\
\text { Given } y=f^{-1}(2 x+1) \\
\Rightarrow 2 x+1=f(y) \\
\Rightarrow x=\frac{1}{2}(f(y)-1) .
\end{array}
$$
Thus, the inverse function of $y=f^{-1}(2 x+1)$ is
$$
y=\frac{1}{2}(f(x)-1) \text {. }
$$
Then, $f(x+1)-2=\frac{1}{2}(f(x)-1)$... | 3+\left(\frac{1}{2}\right)^{n-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,582 |
5. Given that $M$ is the midpoint of edge $C_{1} D_{1}$ of the cube $A_{1} B_{1} C_{1} D_{1}-A B C D$, $O$ is the midpoint of $B D_{1}$, and $O M / /$ plane $\beta$, where plane $\beta$ passes through point $B$ and is different from plane $B_{1} B C C_{1}$. If point $P \in \beta$, and $P$ is within or on the boundary o... | 5. $\sqrt{2}$.
As shown in Figure 2, by the given conditions, point $P$ lies on segment $B C_{1}$. Then the angle $\theta$ formed by $A_{1} P$ and plane $B_{1} B C C_{1}$ is
$$
\begin{array}{l}
\theta=\angle A_{1} P B_{1}, \\
\tan \theta=\frac{A_{1} B_{1}}{B_{1} P} .
\end{array}
$$
Since the minimum value of $B_{1} P... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,583 |
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