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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Four, (1/3 points) There are 99 numbers $a_{1}, a_{2}, a_{3}, \cdots, a_{98}$, defined as: $S_{1} = a_{1}, S_{2} = a_{1} + a_{2}, \cdots, S_{k} = a_{1} + a_{2} + \cdots + a_{k} (1 \leqslant k \leqslant 99)$. There are 100 numbers $b_{1}, b_{2}, \cdots, b_{100}$, and $b_{1} = 1, b_{2} = 1 + a_{1}, b_{3} = 1 + a_{2}, \cd... | $\begin{array}{l}\text { Four, (1) } \because S_{k}^{\prime}=b_{1}+b_{2}+\cdots+b_{k} \\ ==?+\left(1+a_{1}\right)+\left(1+a_{2}\right)+\cdots+\left(1+a_{k-1}\right) \\ =k+\left(a_{0}+a_{2}+\cdots+a_{k-1}\right) \\ =k+S_{k-1} \text {. } \\ \therefore S^{\prime}{ }_{4}-S_{2-1}=k \text {. } \\ \text { (2) } \because A=\fr... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,162 |
Initially 35. 4 distinct prime numbers, none of whose last digit is 1, have a sum of squares equal to \(4pq\) (where \(p, q\) are primes, \(p \neq q\)). \(A=2^n pq\) (where \(n\) is an integer greater than 6) is a 6-digit number. The sum of the two 3-digit numbers formed by the first 3 digits and the last 3 digits of \... | Let
$$
2^{\prime \prime} p q=10^{5} a+10^{4} b+10^{3} c+10^{2} d+10 e+f,
$$
where $a, b, \cdots, f$ are all integers, $1 \leqslant a \leqslant 9,0 \leqslant b, c, \cdots, f \leqslant$
9. Then (1) can be written as
$$
\begin{aligned}
2^{2} p q= & 199(1002+10 s+c)+100(a+d) \\
& +10(b+e)+c+f \\
= & 3^{3} \cdot 37 \cdot 2... | 1995 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,163 |
Initial 36. $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ of $\triangle A B C$ respectively, and $\frac{\operatorname{ctg} B+\operatorname{ctg} C}{\operatorname{ctg} A}=3990$. Find the value of $\frac{b^{2}+c^{2}}{a^{2}}$. | Solving, by the cosine theorem we get
$$
\begin{array}{l}
a^{2}+b^{2}-c^{2}=2 a b \cos C=2 a b \sin C \cdot \operatorname{ctg} C \\
=4 S \operatorname{ctg} C .(S \text { is the area of the triangle) }
\end{array}
$$
Similarly, we get $c^{2}+a^{2}-b^{2}=4 S \operatorname{ctg} B$,
$$
b^{2}+c^{2}-a^{2}=4 S \operatorname{... | \frac{1996}{1995} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,164 |
35. As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at point $P$, with radii $R_{1}$ and $R_{2}$, respectively. Line $M N$ is the external common tangent of $\odot O_{1}$ and $\odot O_{2}$, touching them at points $M$ and $N$, respectively. In $\triangle A B C$, side $B C$ lies on $M N$, ... | Prove: Connect $O_{1} M, O_{1} B, O_{1} E, O_{2} N, O_{2} C, O_{2} F$. Let $\angle M O_{1} B=\alpha, \angle N O_{2} C=\beta$. Then $O_{1} M=R_{1}, O_{2} N=R_{2}$, $\angle O_{1} M B=90^{\circ}, \angle O_{1} E B=90^{\circ}, \angle O_{2} N C=90^{\circ}$, $\angle O_{2} F C=90^{\circ}$. Therefore, $\angle M O_{1} E=2 \alpha... | S_{\triangle A B C} \leqslant 2(3-2 \sqrt{2}) R_{1} R_{2} | Geometry | proof | Yes | Yes | cn_contest | false | 708,165 |
36. As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ intersect at $A, B$. A line through $A$ intersects the two circles at $Y, Z$, and the tangents to the two circles at $Y, Z$ intersect at $X$.
Let the circumcircle of $\triangle O_{1} O_{2} B$ be $\odot O$, and the line $X B$ intersects $\odot O$ at another po... | In $\triangle P Y Z$, we have
$$
\begin{array}{rl}
\angle O_{1} & P O_{2}=\angle Y P Z=180^{\circ} \\
& -\left(\angle O_{1} Y A+\angle O_{2} Z A\right) \\
& =180^{\circ}-\left(\angle O_{1} A Y+\angle O_{2} A Z\right) \\
& =\angle O_{1} A O_{2}=\angle O_{1} B O_{2} .
\end{array}
$$
Therefore, $O_{1}, B, P, O_{2}$ are c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,166 |
Example 8. In a convex quadrilateral $ABCD$, $AC \perp BD$, and let $E$ be the foot of the perpendicular from $A$ to $BD$. Construct the reflections of $E$ about $AB, BC, CD, DA$ as points $P, Q, R, S$ respectively.
Prove that $P, Q, R, S$ are concyclic. | By the property of axial symmetry, we have $A P = A E$
$= A S \Rightarrow$ point $A$ is
the circumcenter of $\triangle E S P \Rightarrow \angle 1$
$=\angle 5$. Similarly, we get
point $B$ is the circumcenter of $\triangle E P Q$
$\Rightarrow \angle 2 = \angle 6$,
point $C$ is the circumcenter of $\triangle E Q R$
$\Rig... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,167 |
28. Let $D, E, F$ be points on the sides $BC, CA, AB$ of $\triangle ABC$, respectively, such that $AD, BE, CF$ all bisect the perimeter of $\triangle ABC$. Prove that: $DE + EF + FD \geqslant \frac{1}{2}(AB + BC + CA)$. | $$
\begin{array}{l}
\text { Prove } \because B D+C D+C E \\
= C D+C E+A E \\
= \text { the semi-perimeter of } \triangle A B C, \\
\therefore \quad A E=B D .
\end{array}
$$
Similarly, $A F=C D$,
$$
B F=C E \text {. }
$$
Let $B C=a, C A=b, A B=c$. Then we have
$$
A E+A F=a, B D+B F=b, C D+C E=c \text {. }
$$
As show... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,168 |
Example 9. In $\triangle ABC$, $AB=AC$, altitude $AH=BC$, extend $AB$ to $E$, such that $BE=AB$, draw $EF \perp AC$, with $F$ as the foot of the perpendicular.
Prove: $\triangle EBH \sim \triangle EHF \sim \triangle HCF$.
---
The original text has been translated into English while preserving the original formatting ... | Draw $E K \perp$
line $A H, K$ is the foot
of the perpendicular. Since $\angle A K E=90^{\circ}$
$$
=\angle A F E \Rightarrow A, E, K \text {, }
$$
$F$ are concyclic.
$\because A K$ bisects $\angle E A F$,
$\therefore \overparen{K E}=\overparen{K F} \Rightarrow K E=K F$ (not drawn).
It is easy to prove $K H=A H=B C=2 \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,169 |
Example 10. On the same straight line, take $A B=B C=C D$, and construct $\odot O$ with $B C$ as the diameter. Draw a tangent line $A E$ from $A$ to $\odot O$, with the point of tangency at $M$.
Prove: $\angle A M B=\angle E M D$.
---
Please note that the translation preserves the original formatting and structure of... | Analyzing the circle $\odot O$ with diameter $M F$, forming $\mathrm{Rt}$ $\triangle A M F$, the ray $M B$ intersects $A F$ at $N$. Since $A O$ is the median of $\triangle A M F$, and point $B$ lies on $A O$ with $A B: B O=2: 1$, point $B$ must be the centroid of $\triangle A M F$. Therefore, $M N$ is another median of... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,170 |
Example 11. In an acute triangle $\triangle ABC$, $\angle C=45^{\circ}$, and $O$ is the circumcenter. A circle passing through points $A, B, O$ intersects $AC, BC$ at $E$, $F$.
Prove: $CO \perp EF, CO=EF$. | Analyze: Connect $EO, FO$ and extend them to intersect $BC, AC$ at $M, N$, then connect $OB$.
By $\angle BOM = \angle A = \frac{1}{2} \angle BOC'$ $\Rightarrow \angle BOM = \angle COM \Rightarrow OM \perp BC$. That is, $EM \perp FC$. Similarly, $FN \perp EC$, so point $O$ must be the orthocenter of $\triangle CEF$, whi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,171 |
Example 12. In convex quadrilateral $A B C D$, $A B=A C$ $=B D, \angle C A B=36^{\circ}, \angle D B A=48^{\circ}$.
Find: $\angle B C D, \angle A D C$. | Analyze: Draw $B H \perp A D$
at $H$ intersecting $A C$ at $S$, connect $D S$, extend $B C$ to $E$. It is easy to know that
$$
\begin{array}{l}
\angle A B H=\angle D B H= \\
\angle D B C=24^{\circ}, \angle A S H= \\
\angle C A B+\angle A B H=36^{\circ}+ \\
24^{\circ}=60^{\circ} \text{. Thus } \angle D S H=\angle A S ... | 126^{\circ}, 96^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,172 |
Example 13. As shown in the figure, $ABCD, PQRS, DQEF$, CSGH are four squares, where
$$
\begin{array}{l}
PR \perp AB, PR= \\
\frac{1}{2} AB .
\end{array}
$$
Prove: $E, R$,
$G$ are collinear, and
$$
ER=RG
$$ | This problem is found in the famous book "Review and Research of Elementary Mathematics (Plane Geometry)". This problem may seem daunting at first, but once the hidden conditions are uncovered, it becomes much easier. Please see: $\square$
From $P R \perp A B, P R=\frac{1}{2} A B \Rightarrow Q S / / D C, Q S$ $=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,173 |
4. On the sides $BC, CD$ of a unit square $ABCD$, there are points $M, N$ respectively, such that $\angle MAN=45^{\circ}$. Prove: (1) $CM \cdot CN=2 BM \cdot DN$;
(2) $\frac{AB+BM}{AD+DN}=\frac{AB^2+BM^2}{AD^2+DN^2}$. | (The key step is to let $B M=m, D N=$ $n$, and try to derive the equation: $m+n+m n=1$ ) | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,179 |
5. $A B C D$ is a trapezoid circumscribed around $\odot O$, where $D C$ // $AB$.
Prove: (1) $A D \cdot B C > A B \cdot D C$;
(2) $A D^{2} + B C^{2} < A B^{2} + D C^{2}$;
(3) $\frac{1}{A \bar{D}} + \frac{1}{B C} < \frac{1}{1 B} + \frac{1}{1 D}$ | (Tian Yi: Let $A B, D C$ be tangent to $\odot \bigcirc$ at $M, N$. Denote $A M:=a, B M:=b, C N:=c, D N:=d$. Try to derive the equal area relationship: $a d=b c$) | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,180 |
6. In $\triangle A B C$, $\angle C=90^{\circ}$. On $A B$, take $A F = A C$, $B E = B C$. A circle is drawn through points $A, C, E$, and another circle is drawn through points $B, C, F$. The other intersection point of the two circles is $K$. Find $\frac{C K}{E F}=$ ? | (Hint: Try to show that point $K$ is the incenter of $\triangle A B C$) | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,181 |
Theorem 1 Suppose we are given two polynomials with coefficients in $K$
$$
\begin{array}{l}
f(x)=a_{m} x^{m}+a_{m-1} x^{m-1}+\cdots+a_{1} x+a_{0}, \\
\varphi(x)=b_{n} x^{n}+b_{n-1} x^{n-1}+\cdots+b_{1} x+b_{0},
\end{array}
$$
where $a_{m} \neq 0, b_{n} \neq 0$. Then there exist polynomials $q(x)$ and $r(x)$ with coeff... | Prove (by induction on the degree $m$ of $f(x)$) that if $m < n$, then we can take
$$
q(x)=0, r(x)=f(x) .
$$
For the case $m \geqslant n$, the polynomial
$$
f_{1}(x)=f(x)-\frac{a_{m}}{b_{n}} x^{m-n} \varphi(x)
$$
has a degree lower than $m$. We can assume there exist polynomials $q_{1}(x)$ and $r_{1}(x)$ with coeffic... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,182 |
Lemma Let $f, g$ and $h$ be polynomials with integer coefficients, and $h(x)=f(x) g(x)$. If a prime $p$ divides all the coefficients of the polynomial $h$, then at least one of the polynomials $f$ and $g$, all of its coefficients are also divisible by $p$. | Proof by contradiction, assume $p \nmid a_{i}, p \nmid b_{j}$, and $i$ and $j$ are the smallest indices satisfying this condition. Then $p \nmid a_{i} b_{j}$. Consider the coefficient of $h$:
$$
c_{i+j}=a_{i+j} b_{0}+\cdots+a_{i} b_{j}+\cdots+a_{0} b_{i+j},
$$
In the right-hand side of this expression, all terms excep... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,183 |
Theorem 3 If the integer-coefficient polynomial $f(x)$ is reducible over the field of rational numbers, then it is also reducible over the ring of integers. | Proof: Let the integer-coefficient polynomial $f(c)$ be factored into a product of rational-coefficient polynomials of lower degree:
$$
f(x) = (x) \psi(x) .
$$
First, bring the coefficients of $\varphi(x)$ to a common denominator, then factor out this common denominator, and subsequently factor out the greatest common... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,184 |
Theorem 5 (Generalization of Eisenstein's Criterion) Let
$$
f(x)=c_{n} x^{n}+c_{n-1} x^{n-1}+\cdots+c_{0}
$$
be a polynomial with integer coefficients. Suppose there exists a prime $p$ and a natural number $m \leqslant n$, such that
(i) $p \nmid c_{n}$,
(ii) $p \mid c_{j}(j=0,1, \cdots, m-1)$,
(iii) $p^{2} \nmid c_{0}... | To prove the following statement: If $f$ can be factored into the product of two integer-coefficient polynomials $\varphi$ and $\psi$
\[
f(x) = \varphi(x) \psi(x),
\]
then one of $\varphi$ and $\psi$ has a degree $\geqslant m$, and this polynomial (for the same $p$ and $m$) satisfies similar conditions (i), (ii), and (... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,187 |
Example 2. (IMO-34-1) Let $n$ be an integer greater than 1. Prove that $f(x)=x^{n}+5 x^{n-1}+3$ is irreducible over the ring of polynomials with integer coefficients.
保持了原文的换行和格式。 | Prove that if $f(x)$ is reducible over the ring of integer polynomials, then according to the generalization of Eisenstein's criterion, we have
$$
\begin{aligned}
f(x)= & \left(x^{n-1}+\alpha_{n-2} x^{n-2}+\cdots+\alpha_{1} x+\alpha\right) \\
& \cdot(x+\beta) .
\end{aligned}
$$
By comparing coefficients, we know
$$
\a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,189 |
Theorem 6 Let $\xi$ be a given complex number. Then, within the specified range of coefficients, the minimal polynomial $f(x)$ of $\xi$ can divide any annihilating polynomial $g(x)$ of $\xi$. | Proof First, perform the division algorithm
$$
g(x)=q(x) f(x)+r(x) .
$$
Here, the remainder $r$ also satisfies the condition $r(\xi)=0$. But the degree of $r$ is less than that of $f$, so $r$ must be the zero polynomial. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,190 |
Theorem 7 Let $\xi$ be a given complex number, and $f$ be the minimal polynomial of $\xi$ with the leading coefficient equal to 1. In the specified coefficient range, $f$ is the minimal polynomial of $\xi$ if and only if $f$ is irreducible. | Proof of Necessity: Suppose $f$ has a non-trivial factorization in the specified coefficient range
$$
f(x)=\varphi(x) \psi(x) .
$$
Then by $\varphi(\xi) \psi(\xi)=f(\xi)=0$,
we can get $\varphi(\xi)=0$ or $\psi(\xi)=0$. Therefore, $f$ is not the minimal polynomial.
Sufficiency: If $f$ is not the polynomial of the low... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,191 |
Example 3. (IMO-36-6) Let $p$ be an odd prime, and consider the subsets $A$ of the set $\{1,2, \cdots, 2 p\}$ that satisfy the following two conditions:
(i) $A$ has exactly $p$ elements;
(ii) The sum of all elements in $A$ is divisible by $p$.
Determine the number of such subsets $A$. | Let $S=\{1,2, \cdots, 2 p\}$.
We take $p$-th roots of unity $\xi=e^{i \frac{i x}{p}}$, and then consider the polynomial
$$
\prod_{j=1}^{2 p}\left(x-\xi^{j}\right)=\left(x^{p}-1\right)^{2}=x^{2 p}-2 x^{p}+1 .
$$
Comparing the coefficient of $x^{p}$, we get:
(1) $2=\sum_{\left\{i_{1}, \cdots, i_{p}\right\} \subset S} \x... | \frac{1}{p}\left(C_{2 p}^{p}-2\right)+2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,192 |
If these roots are also roots of another polynomial $g(x)$, then $f(x) \mid g(x)$.
If these roots are also roots of another polynomial $g(x)$, then $f(x) \mid g(x)$. | To prove the division with remainder:
$$
g(x) = q(x) f(x) + r(x).
$$
Then all the roots of $f(x)$ are also roots of $r(x)$. However, a non-zero polynomial of degree less than $n$ cannot have $n$ roots, so $r$ must be the zero polynomial.
Referring to Example 3, we consider the polynomials
$$
\begin{array}{l}
f(x) = 1 ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,193 |
Example 5. Prove: There do not exist four points $A, B, C, D$ in the plane such that $\triangle A B C, \triangle B C D, \triangle C D A$, $\triangle D A B$ are all acute triangles. | Proof Consider two cases:
(I) The given four points are the vertices of a convex quadrilateral. Since
$$
\begin{array}{l}
\angle A B C+\angle B C D+\angle C D A+\angle D A B \\
=360^{\circ},
\end{array}
$$
$\mathcal{F}=\frac{360^{\circ}}{4}=90^{\circ}$. Without loss of generality, assume $\angle D A B \geqslant 90^{\ci... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,196 |
Second Pump Find all real numbers $p$ such that the cubic equation
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p
$$
has three roots that are all natural numbers. | The original problem is equivalent to
$$
5 x^{2}-5 p x+66 p-1=0
$$
having both roots as natural numbers.
Method 1: Let $u, v$ be the roots of equation (2), then
$$
\left\{\begin{array}{l}
u+v=p, \\
u v=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$
Eliminating the parameter $p$ yields
$$
\begin{array}{l}
5 u v=66 u+66 v... | 76 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,202 |
Third Question As shown in the figure, the incircle $\odot O$ of quadrilateral $ABCD$ touches the sides at $E, F, G$, and $H$. On $\overparen{EF}$ and $\overparen{GH}$, draw tangents to $\odot O$ that intersect $AB$ at $M$, $BC$ at $N$, $CD$ at $P$, and $DA$ at $Q$. Prove: $MQ \parallel NP$. | Method 1 The symmetry of the given figure in this problem easily leads one to think of the analytical method. A Cartesian coordinate system can be established with point $O$ as the origin and the two diagonals of the rhombus as the coordinate axes. The key is the reasonable setting of parameters.
Let $\odot O$ be the ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,203 |
Example 1. Let $0<a<b$, draw lines $l$ and $m$ through the fixed points $A(a, 0), B(b, 0)$ respectively, intersecting the parabola $y^{2}=x$ at two distinct points. When these four points are concyclic, find the locus of the intersection point $P$ of $l$ and $m$. | Let $P\left(x^{\prime}, y^{\prime}\right)$, then the equations of $l$ and $m$ are:
$$
\begin{array}{l}
\quad l: y^{\prime} x-\left(x^{\prime}-a\right) y \\
+a y^{\prime}=0, \\
\quad m: y^{\prime} x-\left(x^{\prime}-b\right) y \\
+b y^{\prime}=0 .
\end{array}
$$
And $f: y^{2}-x=0$, so by Theorem 2 we get
$$
y^{\prime}\... | x^{\prime}=\frac{a+b}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,205 |
Example 6. There are four points $A, B, C, D$ on a plane, where no three points are collinear. Connecting each pair of points results in six line segments. Prove: There must exist a point from which three line segments can form a triangle. | Proof: In the six connected line segments, there must be one that is the longest. Let's assume it is $AB$. Since
$$
\begin{array}{r}
D A+D B> \\
A B, C A+C B>
\end{array}
$$
$A B$,
$$
\begin{aligned}
\quad \therefore D A+D B \\
+C A+C B_{4}> \\
2 A B,
\end{aligned}
$$
or $(D A+C A)+(D B+C B)>2 A B$.
According to Propo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,207 |
Question: Let the function $f_{0}(x)=|x|, f_{1}(x)=\left|f_{0}(x)-1\right|$, $f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed figure formed by the graph of the function $y=f_{2}(x)$ and the $x$-axis is $\qquad$ | Let $S_{n}$ denote the corresponding area, then from Figure 1, it is easy to see:
$$
\begin{array}{l}
S_{0}=0, S_{1}=1, \\
S_{2}=7 .
\end{array}
$$
Let $f_{0}(x)=|x|$,
$$
f_{n}(x)=\left|f_{n-1}(x)-n\right|
$$
$(n \in N)$, and $S_{n}$ represents the area of the closed figure formed by the graph of $f_{n}(x)$ and the $... | S_{n}=\frac{1}{4}\left(2 n^{3}+3 n^{2}-\frac{1+(-1)^{n+1}}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,208 |
1(1956, USA). On the sides of $\triangle A B C$, construct equilateral triangles $B P C, C Q A, A R B$ outwardly, such that $\angle B P C=\angle C Q A = \angle A R B=120^{\circ}$. Then $\triangle P Q R$ is an equilateral triangle.
2(1957, IMO). On the sides of $\triangle A B C$, construct $\triangle B P C, \triangle C ... | Prove: As shown in the figure, let $\angle B P C=\alpha, \angle C Q A=\beta, \angle A R B=\gamma$. Then $\alpha+\beta+\gamma=360^{\circ}$. Therefore, $\angle Q A R+\angle R B P+\angle P C Q=720^{\circ}-(\alpha+\beta+\gamma)=360^{\circ}$.
Since in the isosceles $\triangle A R B$, $R A=R B$, we can rotate $\triangle P B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,209 |
Theorem The sum of the products of the two pairs of opposite sides of a cyclic quadrilateral is equal to the product of the diagonals. | Proof: As shown in the figure, with $D$ as the pole and ray $D O$ as the polar axis, we establish a polar coordinate system. Without loss of generality, let the diameter of $\odot O$ be 1. Then the equation is $\rho=\cos \theta$. Since $A\left(\rho_{1}, \theta_{1}\right), B\left(\rho_{2}, \theta_{2}\right), C\left(\rho... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,210 |
Theorem From a point $P$ in the plane of $\triangle A B C$, draw rays making equal angles $\theta$ with the sides of the triangle in the same direction, intersecting at points $A_{1}, B_{1}$, $C_{1}$. Let the circumradius of $\triangle A B C$ be $R$, and $O P=d$. Then
$$
S_{\triangle A_{1} B_{1} c_{1}}=\frac{\left|R^{2... | Prove that, as shown in the figure, $\angle P A_{1} B=\angle P B_{1} C=\angle P C_{1} A=\theta$. Let $C I$ intersect $\odot O$ at $D$, and connect $A D$, $A P$. Since $P, A_{1}, C, B_{1}$ are concyclic, by the Law of Sines, we have
$$
A_{1} B_{1}=\frac{P C \cdot \sin C}{\sin \theta} .
$$
Similarly, $B_{1} C_{1}=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,211 |
2. As shown in the figure, on the sides $B C, C A$, and $A B$ of the equilateral $\triangle A H e$, there are points $D, E, F$ that divide the sides in the ratio 2 : ($n-2$) (where $n>4$). The segments $A D, B E, C F$ intersect to form $\triangle P Q R$, whose area is $\frac{1}{7}$ of the area of $\triangle A B C$. Wha... | 2. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,215 |
3. If $\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=k$, then the graph of the line $y=k x+k$ must pass through ( ).
(A) the first, second, and third quadrants
(B) the second and third quadrants
(C) the second, third, and fourth quadrants
(D) none of the above | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,216 |
6. The number of triangles that can be formed with the prime factors of 1995 as side lengths is ( ).
(A) 4
(B) 7
(C) 13
(D) 50 | 6. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,220 |
2. In the rectangular paper piece $A B C D$, $A B=6, B C=8$. If the paper is folded so that $A$ coincides with $C$, then the length of the fold line $E F$ is | 2. According to the problem, the fold line $E F$ is the perpendicular bisector of $A C$. Let $A F=x$, then $F C=x$. Since $A B$ $=6, B C=8$, so $A C=10, A O=O C=5$. In
Rt $\triangle C D F$, $6^{2}+(8-x)^{2}=x^{2}$, we get $x=\frac{25}{4}$. In Rt $\triangle A O F$, $\left(\frac{25}{4}\right)^{2}-5^{2}=O F^{2}, O F^{2}=... | \frac{15}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,222 |
3. In $\triangle A B C$, $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively. $a=15, b=17, \angle A=\theta(\theta$ is a constant). If the $\angle C$ of the triangle satisfying the above conditions is unique, then $\operatorname{tg} C=$ $\qquad$ | $\angle A=g$ is a constant, $a<b, \angle A$ must be an acute angle. Given that $\angle C$ has a unique value satisfying the conditions of the problem, thus $\angle E$ $=180^{\circ}-\angle C-\theta$ is also unique, which means $\angle B=90^{\circ}$. Therefore, $c=\sqrt{17^{2}-15^{2}}=8, \cos C=\frac{8}{15}$. | \frac{8}{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,223 |
4. If the prime numbers $m, n$ satisfy $5 m+7 n=129$, then the value of $m+n$ is $\qquad$ . | \&. The original wing converts to $5 m \equiv 129(\bmod 7)$, i.e., $5 m \equiv 3$ (modi), solving yields $m \equiv 2(\bmod 7)$. Given that $m$ is a prime number, and from the original indeterminate equation, $m<25$. Therefore, the possible values of $\boldsymbol{m}$ are 2 or 23. Checking, when $m=2$, $n=17$, thus, $m+n... | 19 \text{ or } 25 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,224 |
5. Given $x+\frac{1}{x}=3$, then $x^{4}+3 x^{3}-16 x^{2}+3 x-1$ ? The value equals $\qquad$ . | 5. From $x^{2}+1=3 x$, we get $x^{4}=7 x^{2}-1, x^{3}=3 x^{2}-x$. Therefore, the original expression $=7 x^{2}-1+9 x^{2}-3 x-16 x^{2}+3 x-17$ $=-18$.
Another solution: From the given $x^{2}+1=3 x, x^{2}-3 x=-1$, so,
$$
\begin{array}{l}
x^{4}+3 x^{3}-16 x^{2}+3 x-17=\left(x^{4}+2 x^{2}+1\right)+3 x^{3}+3 x \\
-18 x^{2}-... | -18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,225 |
6. Let $x_{1}, x_{2}, \cdots, x_{19}$ be integers, and satisfy $x_{1}+x_{2}+\cdots$ $+x_{19}=95$. Then the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{19}^{2}$ is $\qquad$. | 6. Let's assume $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant \cdots \leqslant x_{19}$. Note that when $1<x_{i}$ $\leqslant x_{j}$, $x_{i}^{2}+x_{j}^{2}<x_{i}^{2}+1-2 x_{i}+2 x_{j}+1+x_{j}^{2}=\left(x_{i}-\right.$ $1)^{2}+(x,+1 \cdot 1)^{2}$. Therefore, when $x_{1}=x_{2}=\cdots=x_{18}=1, x_{10}=$ 77, $x^{2} \cdot x_... | 5947 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,226 |
One, (This question is worth 20 points)
In the border desert area, patrol vehicles travel 200 kilometers every day. Each vehicle departs from base $A$ at the same time, completes the mission, and then returns along the original route to the base (returning together). After vehicles A and B reach point $B$ on the way, t... | When the car reaches point $B$ on the way, it takes $x$ days, and from $B$ to the farthest point, it takes $y$ days, then we have
$$
2[3(x+y)+2 x]=14 \times 5,
$$
which simplifies to $5 x+3 y=35$.
According to the problem, we need $x>0, y>0$, and
$$
14 \times 5-(5+2) x \leqslant 14 \times 3 \text {, }
$$
which simpli... | 1800 | Other | math-word-problem | Yes | Yes | cn_contest | false | 708,227 |
II. (Full marks for this question:
20 points)
Through a point $P$ outside $\odot O$
draw a tangent $P_{N}$ to $\odot O$,
with the point of tangency being $N$. Let $M$ be the midpoint of $P N$. Then the circle passing through $P$ and $M$
intersects $\odot O$ at $A$,
$B$, and the extension of $B A$ intersects $P N$ at ... | $$
\text { II. QM } Q Q P=Q A \cdot Q B=Q N^{2} \text {, }
$$
get $\frac{Q M}{Q N}=\frac{Q N}{Q P}$.
Let $Q . A=x, Q N=y$, then $M N=M P=x+y$.
Thus, $P=x+(x+y)=2 x+y$.
From $\wedge(*), \frac{x}{y}=\frac{y}{2 x+y}$.
Rearranging gives $(x+y)(2 x-y)=0$.
Therefore, $y \approx 2 x$, or $x=-y$ (not valid).
$E_{1} . M T V=P ... | P M=3 M Q | Geometry | proof | Yes | Yes | cn_contest | false | 708,228 |
1. Given, on the midline $M N$ of $\angle A B C$, take any point $P$, the rays $B P, C P$ intersect $A C, A B$ at $F, E$ respectively.
Try to prove: $\frac{A E}{E B}+\frac{A F}{F C}=1$. | Draw a line through $A$ parallel to $B C$ intersecting the rays $B F$, $C E$ at $B^{\prime}, C^{\prime}$. Clearly, $\triangle P B^{\prime} C^{\prime} \sim \triangle P B C$, and we have
$$
\begin{array}{l}
\frac{A E}{E B}=\frac{A C^{\prime}}{B C}, \\
\frac{A F}{F C}=\frac{A B^{\prime}}{B C} .
\end{array}
$$
Therefore, ... | 1 | Geometry | proof | Yes | Yes | cn_contest | false | 708,229 |
Three, (This question is worth 20 points)
Find the real number pairs $(p, q)$ such that the inequality $\left|x^{2}+p x+q\right| \leqslant 2$ holds for all $1 \leqslant x \leqslant 5$.
untranslated part:
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。
---
Three, (This question is worth 20 points)
Find the real number pairs $(p,... | Let $f(x)=x^{2}+p x+q$, the axis of symmetry of the graph of this quadratic function is $x=-\frac{p}{2}$, and it opens upwards.
(1) When $-\frac{P}{2} \leqslant 1$, the interval $[1,5]$ is to the right of the axis of symmetry, so $-2 \leqslant f(1)5$ when, there is $-2 \leqslant f(5)<f(1) \leqslant 2$, i.e.,
$$
\left\{... | (-6,7) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,230 |
2. The average of 30 odd prime numbers is closest to ( ).
(A) 12
(B) 13
(C) 14
(D) 15 | 2. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,232 |
3. A pentagram has five vertices. Draw straight lines through these five vertices so that each line must pass through two of them. Then, the total number of lines that can be drawn is ( ).
(A) 5
(B) 10
(C) 15
(D) 20 | 3. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 708,233 |
4. For a product, the selling price after a reduction of $a$ yuan is 85% of the original price, then the original price of this product is ( ).
(A) $(1+85 \%) a$ yuan
(B) $(1-85 \%) a$ yuan
(C) $\frac{a}{1-85 \%}$ yuan
(D) $\frac{a}{1+85 \%}$ yuan | 4. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,234 |
5. As shown in the figure, $\angle A O C$ is a right angle, $\angle C O D=21.5^{\circ}$, and $O B$, $O D$ are the bisectors of $\angle A O C$, $\angle B O E$ respectively. Then $\angle A O E$ equals ( ).
(A) $111.5^{\circ}$
(B) $133^{\circ}$
(C) $134.5^{\circ}$
(D) $178^{\circ}$ | 5. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,235 |
6. If $a, b, c$ are all positive integers, and $a, b$ are odd, then $3^{a}+(b-1)^{2} c$ is ( ).
(A) only when $c$ is odd, its value is odd
(B) only when $c$ is even, its value is odd
(C) only when $c$ is a multiple of 3, its value is odd
(D) regardless of $c$ being any positive integer, its value is always odd | 6. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,236 |
7. In the final exam, Wang Fang's scores in Chinese, Mathematics, English, and Politics are as follows: The sum of the scores in Mathematics and Politics equals the sum of the scores in Chinese and English, but the sum of the scores in Chinese and Mathematics exceeds the sum of the scores in English and Politics, and t... | 7. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 708,237 |
Example 2. As shown in the figure, semicircle $O$ is tangent to the sides $AD$, $DC$, and $BC$ of quadrilateral $ABCD$, with center $O$ on side $AB$. If points $A, B, C, D$ lie on a circle, then $AB = AD + BC$. (23rd IMO problem) | Extend
$A D, B C$ to meet at point
$P$, draw $D^{\prime} C^{\prime} \parallel$
$A B$, such that $D^{\prime} C^{\prime}$ is
tangent to the semicircle, $D^{\prime} \in$
$P A, C^{\prime} \in P B$. It is easy to
see that $\angle P D^{\prime} C^{\prime}=$
$\angle A=\angle P C D \Rightarrow \triangle P D^{\prime} C^{\prime} ... | AB = AD + BC | Geometry | proof | Yes | Yes | cn_contest | false | 708,240 |
6. A certain two-digit natural number can be divided by the sum of its digits, and the quotient is exactly a multiple of 7. Write down all the two-digit numbers that meet the conditions: $\qquad$ . | $6.21,42,63,84$,
| 21, 42, 63, 84 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,244 |
Three, (11 points) There are four numbers. Each time, select three of them, calculate their average, and then add the other number. Using this method, four calculations were made, resulting in the following four numbers: 86, 92, 100, 106. (1) Find the four numbers, (2) Find the average of these four numbers. | Three, let these four numbers be $a, b, c, d$. We can get
$$
\left\{\begin{array}{l}
\frac{a+b+c}{3}+d=86, \\
\frac{b+c+d}{3}+a=92, \\
\frac{c+d+a}{3}+b=100, \\
\frac{d+a+b}{3}+c=106 .
\end{array}\right.
$$
Solving it, we get $a=42, b=54, c=63, d=33$.
$$
\frac{1}{4}(a+b+c+d)=48 \text{. }
$$
Answer: These four numbers... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,246 |
Four, (11 points) A five-digit number, if the sum of the three-digit number represented by the first three digits and the two-digit number represented by the last two digits can be divided by 11, determine whether this five-digit number can be divided by 11, and explain the reason. | Let the first three digits be $x$, and the last two digits be $y$. Then this five-digit number can be expressed as $100x + y$. Since $x + y$ is divisible by 11, we can set $x + y = 11n$ ($n$ is a positive integer). And $100x + y = 99x + (x + y) = 99x + 11n = 11(9x + n)$, so this five-digit number is divisible by 11. | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,247 |
3. If $a, b, c$ are all positive integers, and $a, b$ are odd. Then $3^{a}+(b-1)^{2} c$ is ( ).
(A) only when $c$ is odd, its value is odd
(B) only when $c$ is even, its value is odd
(C) only when $c$ is a multiple of 3, its value is odd
(D) regardless of $c$ being any positive integer, its value is always odd | 3. D
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,250 |
Example 3. In the square $ABCD$, there is a point $P$. If $PA, PB, PC, PD$ divide $\triangle ABCD$ into four triangles with areas in geometric progression, try to determine: How is the position of point $P$ determined? | Analysis: Draw $EF \perp AB$ through $P$, with $E, F$ as the feet of the perpendiculars. It is easy to prove that $S_{\triangle PAB} + S_{\triangle PCD} = \frac{1}{2} \cdot EF \cdot AB = \frac{1}{2} S_{DABCD}$. Therefore,
$S_{\triangle PAB} + S_{\triangle PCD} = S_{\triangle PBC} + S_{\triangle PDA}$.
Let the areas of... | P \text{ is the intersection point of the diagonals of } \square ABCD | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,251 |
4. The possible positive integer values of $k$ that make the equation $k x-12=3 k$ have integer solutions for $x$ are $\qquad$ . | 4. $1,2,3,4,6,12$ | 1,2,3,4,6,12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,258 |
Example 1. Numbers $x\left(x \neq 0^{\circ}\right)$ and $y$ are such that for any $n \geqslant 1$, the number $\overline{x \cdots x 6 y \cdots y}$ is a perfect square. Find such $x$ and $y$. (18th All-Union Mathematical Olympiad for Secondary Schools) | Given the condition $\overline{x 6 y 4}=m^{2}$, since the hundreds digit is 6 and the units digit is 4, we have $40<n<100$, and the last digit of this two-digit number must be 2 or 8. Since $68^{2}=4624$ and $98^{2}=9604$, $m$ can only be 68 or 98.
Thus, we conjecture:
(1) $4 \cdots 462 \cdots 24=\overline{6 \cdots 68^... | x=4, y=2 \text { or } x=9, y=0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,262 |
Example 2. $C$ is a moving point outside the fixed line $AB$. Construct squares $CADF$ and $CBEG$ outwardly on $AC$ and $BC$, respectively. Regardless of how the position of point $C$ changes (on the same side of $AB$). Prove:
(1) The sum of the distances $D D^{\prime}$, $E E^{\prime}$ from $D$ and $E$ to the line $AB$... | Analyzing Figure 1-2, when $C$ moves from a general position to a special position, i.e., $C$ is equidistant from $A$ and $B$, and $\angle A C E=90^{\circ}$, drawing $C C^{\prime} \perp A B$, it is not difficult to find:
(1) $C C^{\prime}$: the midline of the rectangle;
(2) $C C^{\prime}=\frac{1}{2} A B=$ a constant.
T... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,263 |
Example 11. There are 100 points on a plane, where the distance between any two points is no less than 3. Now, connect every two points whose distance is exactly 3 with a line segment. Prove: such line segments will not exceed 300.
(1984, Beijing Junior High School Mathematics Competition) | Given that the central angle subtended by an arc between two points is less than $60^{\circ}$, it means the distance between these two points is less than 3, which contradicts the problem's condition. Therefore, with $P$ as the endpoint, the maximum number of line segments that can be drawn to $A_{1}, A_{2}, A_{3}, A_{... | 300 | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,264 |
Example 1. Find $\frac{1}{1-\frac{1}{1-\cdots \frac{1}{1-\frac{355}{113}}}}$. | Let $a_{-}=\underbrace{-\frac{1}{1-\frac{1}{1-\cdots \frac{1}{1-\frac{355}{113}}}}}_{m}$.
$a_{1}=\frac{1}{1-\frac{355}{113}}=-\frac{113}{242}, a_{2}=\frac{244}{355}, a_{3}=\frac{355}{113}$,
$a_{4}=-\frac{113}{242}, a_{5}=\frac{242}{355}, \cdots$. From this, it is easy to see that $\left\{a_{n}\right\}$ is a periodic se... | \frac{355}{113} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,265 |
Example 2. Find all natural numbers with the following property: its unit digit is 6, and when its unit digit is moved to the front of the number, the resulting new number is four times the original number. | Let the required number be $a_{n} a_{n-1} \cdots \cdots a_{2} a_{1} 6$. According to the problem, we have $\overline{6 a_{n} a_{n-1} \cdots a_{2} a_{1}}=4 \overline{a_{n} a_{n-1} \cdots a_{2} a_{1} 6}$, where $0 \leqslant a_{i} \leqslant 9, a_{i} \in N$.
That is 6
$$
\begin{array}{l}
6 \cdot 10^{n}+a_{n} \cdot 10^{n-1}... | 153846 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,266 |
Example 3. Let $S$ be a subset of the set of numbers $\{1,2,3, \cdots, 1989\}$, and the difference between any two numbers in $S$ is not equal to 4 or 7. How many numbers can $S$ contain at most? | Analysis Notice that the sum of 4 and 7 is 11. We first observe the first 11 natural numbers $1,2, \cdots, 11$, and find that $1,4,6,7,9$ satisfy the conditions of the problem. Then we observe the 11 numbers $12,13, \cdots, 22$, and find that $1+11,4+11,6+11,7+11,9+11$ also satisfy the conditions of the problem. And $1... | 905 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,267 |
Example 4. Prove that in the sequence with the general term $a_{n}=1+2^{2}+3^{3}+\cdots+$ $n^{n}$, there are infinitely many odd composite numbers.
(20th Soviet High School Mathematics Competition) | We first examine the sequence of remainders when $n^{*}(n \in N)$ is divided by 3: $1,1,0,1,2,0,1,1,0,1,2,0, \cdots$. It is not hard to see that it is a periodic sequence with a period of 6.
Therefore, for any $i+E N, \sum_{i=k}^{t+s} i^{i} \equiv 2(\bmod 3)$. Thus, $\sum_{i=1}^{k+35} i^{i} \equiv 0(\bmod 3)$. Consequ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,268 |
Example 5. Given a sequence of natural numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, where $a_{1}$ cannot be divisible by 5, and for each $n$, the equation $a_{n+1} = a_{n} + b_{n}$ holds, where $b_{n}$ is the last digit of $a_{n}$. Prove that the sequence contains infinitely many powers of 2.
(1994, Russian Mathemati... | Since $a_{1}$ cannot be divisible by 5, $b_{1}$ cannot be 0 or 5, thus it is easy to see (by taking $a_{1}$ as $1,2,3,4,6,7,8,9$, etc.) that $b_{2}$ can only be one of $2,4,6,8$. Therefore, the sequence $b_{2}, b_{3}, \cdots$, must be a periodic sequence with a period of 4: $\cdots 2,4,8,6,2,4,8,6, \cdots$. Thus, for a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,269 |
1. Given the arithmetic sequence (1): $5,8,11, \cdots$, and the arithmetic sequence (2): $1,5,9, \cdots$, both having 1996 terms, then there are $\qquad$ numbers that appear in both sequences. | (Answer: 498) | 498 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,271 |
Example 1. Given as shown, $M, N$ are the midpoints of the sides $A D, B C$ of a trapezoid, respectively. On the extension of side $C D$, a point $P$ is taken, and $P M$ intersects the diagonal $A C$ at $Q$. Prove: $N M$ bisects $\angle D N Q$.
保留源文本的换行和格式，直接输出翻译结果如下：
Example 1. Given as shown, $M, N$ are the midpoin... | Prove that $A C$ intersects $M N$ at $O$. It is easy to know that $M O=N O$. Extend $P C, Q N$ to intersect at $R$.
$$
\because M N / / C D, M O=N O \text {, }
$$
By theorem, we have $P C=R C$.
$$
\begin{array}{l}
\because N C \perp P R, \\
\therefore P N=R N . \\
\therefore P M: Q M=R N: Q N=P N: Q N . \\
\therefore ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,279 |
Example 3. In trapezoid $ABCD$, diagonal $AC$ is equal to side $BC$, $M$ is the midpoint of base $AB$, $P$ is a point on the extension of side $DA$, and $PM$ intersects $BD$ at $N$. Prove: $\angle ACP = \angle BCN$.
(1992, Sichuan Province Junior High School Mathematics Competition) | Proof: Let $PC$ intersect $AB$ at $E$, $CN$ intersect $AB$ at $F$, and extend $PN$, $DC$ to intersect at $G$.
$\because$ Through point $P$
$AB, CD$,
$\therefore \frac{AE}{AM}$
$=\frac{CD}{DG}$.
$\because$ Through point
$N$ three lines
intersect $AB, CD$,
$$
\begin{array}{c}
\therefore \frac{BF}{BM}=\frac{CD}{DG}. \\
\b... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,281 |
Column 4. Given as follows:
In $\triangle ABC$, points $D, E$ are on sides $BC$, $AB$ respectively. $AD$, $CQ$ intersect at $F$, $BF$, $DE$ intersect at $G$. Through $G$, a line parallel to $BC$ intersects $AB$, $CE$, $AC$ at $M$, $H$, $N$ respectively. Prove: $GH = NH$. | Proof Let $M N$ intersect $A D$ at $O$, the three lines passing through $E, F, A$ intersect $B C$ at $B, D, C$ respectively. By the Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{B D}{D C}=\frac{M G}{G H}, \\
\frac{B D}{D C}=\frac{G O}{O H}. \\
\frac{B D}{D C}=\frac{M O}{O N}.
\end{array}
$$
From (1) and (2), ap... | GH = HN | Geometry | proof | Yes | Yes | cn_contest | false | 708,282 |
Example 5. Given as shown, $H$ is any point on the internal bisector $A D$ of $\triangle A B C$. $B H, C H$ intersect $A C, A B$ at $E, F$ respectively.
Prove: $\angle E D H = \angle F D H$.
(1985. Tongzhou Middle School Mathematics Competition, 1394, Comprehensive High School Mathematics Competition) | Prove that through $A$ draw the parallel line to $B C$ intersects the extensions of $C F$, $D F$, $D E$, $B E$ at $Q, K, G, P$ respectively. The perpendiculars from $H, E, F$ intersect $B C$ at points $3, D, C$ respectively. Apply the theorem three times:
$$
\begin{array}{l}
\frac{B D}{C D}=\frac{A P}{A \bar{Q}} \\
\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,283 |
Example 4. There are 111 points in an equilateral triangle with a side length of 15. Prove that a circle with a diameter of $\sqrt{3}$ can always cover at least 3 of these points (when covering, part of the coin can be outside the triangle).
| Proof As shown in Figure 3-1, divide each side of an equilateral triangle into ten equal parts, and draw lines parallel to the other two sides through each division point. Then, an equilateral triangle with a side length of 15 is divided into 55 upward (triangles shaped like “$\triangle$”) equilateral triangles with a ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,285 |
Example 7. (Fermat's Theorem) Using line segment $AB$ as the diameter, construct a semicircle on one side, and on the other side construct quadrilateral $ABCD$, such that $AB = \sqrt{2} AD$. Let $P$ be any point on the semicircle, and let $PC, PD$ intersect $AB$ at $E, F$, respectively. Then $AE^2 + BF^2 = AB^2$. | Prove as shown in the figure, extend $P A, P B$ to intersect line $C D$ at $M, N$ respectively. Connect $A N, B M$.
By the Pythagorean theorem, we have
$$
\begin{array}{l}
M C^{2}=M B^{2}-B C^{2}=P M^{2}+P B^{2}-B C^{2}, \\
N D^{2}=N A^{2}-A D^{2}=N P^{2}+P A^{2}-A D^{2} .
\end{array}
$$
(1) + (2) gives
$$
\begin{array... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,286 |
Example 5. Given that $a, b, c$ are distinct real numbers, and $a b c \neq 0$.
Solve the system of equations $\left\{\begin{array}{l}\frac{x}{b^{3}}-\frac{y}{b^{2}}+\frac{z}{b}=1, \\ \frac{x}{c^{3}}-\frac{y}{c^{2}}+\frac{z}{c}=1, \\ \frac{x}{a^{3}}-\frac{y}{a^{2}}+\frac{z}{a}=1 .\end{array}\right.$
(1987, Fujian Provin... | Solving this problem directly is quite challenging. From each equation in the system, we can see that the structure is the same. Here, we might as well treat $x, y, z$ as known quantities and consider $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ as the unknowns. It is easy to see that $\frac{1}{a}$, $\frac{1}{b}$, $\frac{1}... | \left\{\begin{array}{l} x=a b c, \\ y=a b+b c+c a, \\ z=a+b+c \end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,296 |
23. (UK) If the sum of all elements in a finite set of positive integers is a multiple of the least common multiple of all the numbers in the set, then the set is called a "sum-multiple set". Prove that every finite set of positive integers is a subset of some sum-multiple set.
保留源文本的换行和格式，直接输出翻译结果。 | Proof Let the given finite set be $S=\left\{a_{1}, a_{2}, \cdots, a_{r}\right\}$. Let $s=a_{1}+a_{2}+\cdots+a_{r}, m=\left[a_{1}, a_{2}, \cdots, a_{r}\right]$. Suppose $m=2^{k} n$, where $n$ and $k$ are non-negative integers and $n$ is odd. Let the binary expansion of $n$ be
$$
n=\varepsilon_{0}+\varepsilon_{1} 2+\cdot... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 708,297 |
25. (Vietnam) Let $a$ be a known number, $|a|>1$. Solve the system of equations
$$
\left\{\begin{array}{l}
x_{1}^{2}=a x_{2}+1, \\
x_{2}^{2}=a x_{3}+1, \\
\cdots \cdots \\
x_{090}^{2}=a x_{1000}+1, \\
x_{1000}^{2}=a x_{1}+1 .
\end{array}\right.
$$ | Solve for the case when $a1$, so we only need to solve for the case $a>1$.
Since the left side of the equation is non-negative, we have $x_{i} \geqslant-\frac{1}{a}>-1$, $i=1,2, \cdots, 1000$. Also, because the system of equations is cyclically symmetric, we can assume
$$
x_{1}=\max \left\{x_{1}, x_{2}, \cdots, x_{100... | t_{1}=\frac{1}{2}\left(a+\sqrt{a^{2}+4}\right), t_{2}=\frac{1}{2}\left(a-\sqrt{a^{2}+4}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,299 |
Three, fold a rectangular paper strip $A B C D$ along a line $\alpha \neq \frac{3}{4}$
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。
---
Three, fold a rectangular paper strip $A B C D$ along a line $\alpha \neq \frac{3}{4}$ | Fold the short side along $E F$, the insect handle is on $B C$, and $F$ is on $D A$. Clearly, the midpoint of $E F$ is the center $O$ of the rectangle $A B C D$, so $A E C F$ is a rhombus. $C E=A E>B E, \triangle A B E$ has an area less than $\frac{1}{4}$. The area of the pentagon $C D F E B^{*}$ = the area of $\triang... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,303 |
One, prove that in a triangle ABC, there exists a point $P$, such that $P$ is the Fermat point.
The translation is provided while preserving the original text's line breaks and format. | In the acute $\triangle ABC$, arcs $\overparen{AB}$ and $\overparen{AC}$ are constructed on sides $AB$ and $AC$ respectively, with the corresponding inscribed angles being
$$
\begin{array}{l}
240^{\circ}-\angle CAB-\angle ABC, \\
240^{\circ}-\angle BCA-\angle CAB. \\
\overparen{AB} \text{ is denoted as }
\end{array}
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,305 |
II. A sequence's first 5 terms are $1,2,3,4,5$, starting from the 6th term, each term is 1 less than the product of all preceding terms. Prove that the product of the first 70 terms of this sequence is exactly equal to the sum of their squares. | $$
\begin{array}{l}
\text{Let the } n \text{th term be } a_{n}, a_{6}=119. \text{ For } n \geqslant 6, \\
a_{n+1}=a_{1} c_{2} \cdot \cdots \cdot a_{n-1}=a_{n}\left(a_{n}+1\right)-1 \text{.} \\
\text{Therefore, } a_{n+1}-a_{n}+1=a_{n}^{2}. \\
\sum_{n=6}^{70} a_{n}^{2}=\sum_{n=1}^{70} a_{n}^{2}-55 \\
=\sum_{n=6}^{70}\lef... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,306 |
Example 6. What is the largest even integer that cannot be written as the sum of two odd composite numbers? (If a positive integer can be divided by a positive integer other than 1 and itself, then this positive integer is called a composite number).
(2nd American Invitational Mathematics Examination) | Solution: Because the last digit of a positive even number is $0, 2, 4, 6, 8$.
When the last digit is 0, there are even integers such as $0,10,20,30,40, \cdots$, among which $30=15+15$.
When the last digit is 2, there are even integers such as $2,12,22,32,42,52, \cdots$, among which $42=15+27,52=25+27, \cdots$. Theref... | 38 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,307 |
Three, let $A K, B L, C M$ be the angle bisectors of $\triangle A B C$, with $K$ on $B C$. Let $P, Q$ be points on $B L, C M$ respectively, such that $A P = P K, A Q = Q K$. Prove: $\angle P A Q = 90^{\circ} - \frac{1}{2} \angle B A C$. | Three, let $B L$ intersect the circumcircle of $\triangle A B K$ at point $P^{\prime}$, then
$$
\begin{array}{l}
\angle P^{\prime} A K=\angle P^{\prime} Y K \\
=\angle P^{\prime} B . A \cdots \angle P^{\prime} K A \text { because }
\end{array}
$$
this implies $P^{\prime} A=P^{\prime} K$, so $P$ and $P^{\prime}$ are th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,308 |
And, a simple polygon on a plane refers to a closed, non-self-intersecting shape.
1. On the plane, do there exist two congruent simple heptagons, both of which have 7 points as common vertices, but no overlapping sides?
2. Do there exist three simple heptagons that satisfy the above conditions? | 1. Exist.
Let the vertex $A$ of $\triangle A B B^{\prime}$ be on $l$, and $B, B^{\{\prime\}}$ be symmetric about $l$. Take a rectangle $C C^{\prime} D^{\prime} D$ inside $\triangle A B B^{\prime}$, where $C, C^{\prime}$ and $D, D^{\prime}$ are symmetric about $l$. Then the simple heptagons $A B D^{\prime} B^{\prime} C^... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,309 |
Six, in a game played on a $1 \times 1000$ board, initially there are $n$ chips placed. The game is played by two players taking turns. The first player places a chip in any unoccupied cell, and the second player removes any number of chips occupying consecutive cells. If the first player can place all $n$ chips on the... | Six, let the two players be $A$ and $B$.
1. On the board, arrange 96 chips in the pattern of 8 chips, 1 empty space, 8 chips, 1 empty space, $\cdots$. This is a winning position for $A$. Clearly, $A$ can achieve this winning position. After $B$ moves a certain number of consecutive chips, $A$ can return them to their o... | 98 | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,310 |
Three, the first digit of a six-digit number is 5. Is it always possible to add 6 more digits in front of it, so that the resulting twelve-digit number is a perfect square?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Three, assuming it is always possible, $10^{5}$ six-digit numbers starting with 5 can at least generate $10^{5}$ square numbers, then
$$
5 \times 10^{11} \leqslant n^{2}<6 \times 10^{11} \Rightarrow 7 \times 10^{5}<n<8 \times 10^{5} .
$$
There are not $10^{5}$ integers between $7 \times 10^{5}$ and $8 \times 10^{5}$, ... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,312 |
Four, on a plane, there are three different points $A, B, C$. Construct a line $m$ passing through point $C$, such that the product of the distances from points $A, B$ to $m$ is maximized! For each set of points $A, B, C$, is such an $m$ unique? | Let $BC = a, AC = b, \angle BCA = \gamma$. If points $A$ and $B$ are on the same side of line $m$, and let $\theta$ be the angle between $m$ and $CA$, then the angle between $m$ and $CB$ is $\pi - \gamma - \theta$. The product of the distances from points $A$ and $B$ to $m$ is
$$
ab \sin \theta \sin (\pi - \gamma - \th... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,313 |
一、 $P$ is a point on the plane of convex quadrilateral $A B C D$, the angle bisectors of $\angle A P B, \angle B P C, \angle C P D, \angle D P A$ intersect $A B, B C, C D, D A$ at points $K, L, M, N$ respectively.
1. Find a point $P$ such that $K L M N$ is a parallelogram;
2. Determine the locus of all such points $P$. | 1. Let $P$ be the intersection of the perpendicular bisectors of $AC$ and $BD$, then $PA=PC, PB=PD$. Since $PK$ bisects $\angle APE$ and $PL$ bisects $\angle BPC$, we have
\[
\frac{AK}{KB} = \frac{PA}{PB} = \frac{PC}{P} \frac{C}{B} = \frac{CT}{LB}. \text{ Thus } X // AC.
\]
Similarly, $MN // AC$. Therefore, $KK // RN$.... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,314 |
Three, divide a rectangle of size $a \times b(a>b)$ into several right triangles with one common vertex, or no common points. If two triangles have a common side, then this side is the leg of one triangle and the hypotenuse of the other triangle. Prove: $a \geqslant 2 b$. | $O A_{1}, O A_{2}, \cdots, O A_{n}$ are drawn outward. Suppose $O A_{1}$ is a leg of $\triangle O A_{1} A_{2}$, then $O A_{1}$ is the hypotenuse of $\triangle O A_{1} A_{2}$, $O A_{1} > O A_{2}$. $O A_{1}$ is the leg of $\triangle O A_{1} A_{2}$, so $O A_{2}$ is the hypotenuse of $\triangle O A_{2} A_{3}$, $O A_{2} > O... | a \geqslant 2b | Geometry | proof | Yes | Yes | cn_contest | false | 708,316 |
Example 7. Assembling products A, B, and C requires three types of parts, $A, B, C$. Each piece of product A requires 2 $A$ and 2 $B$ parts; each piece of product B requires 1 $B$ and 1 $C$ part; each piece of product C requires 2 $A$ and 1 $C$ part. Using the inventory of $A, B, C$ parts, if $p$ pieces of product A, $... | Prove: No matter how the number of products A, B, and C is changed, it is impossible to use up all the inventory of parts $A, B, C$.
Analyzing this problem, a direct proof is very difficult, so we need to change the proof strategy and use the method of contradiction in indirect proof.
List the conditions and relations... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,318 |
Let $\triangle A B C$ have an area of $1,0<x \leqslant 1, A^{\prime}, B^{\prime}, C^{\prime}$ be points on $B C, C A, A B$ respectively, and $B A^{\prime}: A^{\prime} C=C B^{\prime}: B^{\prime} A$ $\Rightarrow A C^{\prime}=C^{\prime} B=(1-x)+x$. Try to express the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ us... | $\cdots$ 、Obviously, $B A^{\prime}=(1-x) B C, B C^{\prime}=x B A$, therefore
$$
\begin{array}{l}
S_{\triangle C^{\prime}}=\frac{1}{2} B A^{\prime} \cdot B C^{\prime} \sin B \\
=\frac{1}{2}(1-x) x \\
=B A \cdot B C \sin B \\
=x(1-x) \\
\quad \cdot S_{\triangle A B C} \\
=x(1-x) .
\end{array}
$$
I can still deduce
$$
\b... | 3x^2 - 3x + 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,321 |
II. Write the number $1234567802011 \cdots 19941995$ on the blackboard, forming the integer $N_{1}$. Erase the digits of $N_{1}$ that are in even positions, leaving the remaining digits to form the integer $N_{2}$. Remove the digits of $N_{2}$ that are in odd positions, leaving the remaining digits to form the integer ... | $$
=、 9 \times 1+90 \times 2+900 \times 3+996 \times 4=6873 \text {. }
$$
The integer $N_{1}$ is composed of 6873 digits, which are sequentially numbered as $1,2,3, \cdots, 6873$. We still examine the set of numbers $\{1,2, \cdots, 6873\}$. Removing the digits at even positions from $N_{1}$ is equivalent to removing $... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,322 |
Determine all real-coefficient polynomials $p(x)$ such that
$$
t s^{\prime}(t \cdots 1)=(t-2) p(t)
$$
holds for all real numbers $t$. | For $t=2, 2 p(1)=0$, i.e., 1 is a $\cdots$ root of $p(x)$. For $t=1, p(0)=-p(1)=0$. Thus, 0 is a root of $p(x)$. Therefore, we can write $p(x)$ as $x(x-1) q(x)$, where $q(x)$ is a real polynomial. Substituting $p(t)=t(t-1) q(t)$ into the given equation,
\[
\begin{array}{l}
t(t-1)(t-2) q(t-1) \\
=(t-2) t(t-1) q(t), \\
... | p(x)=c x(x-1), c \in \mathbb{R} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,324 |
Item, let's consider a right-angled triangle that satisfies the following conditions:
1. The length of one of the legs of $\triangle$ is an integer 3
2. If the perimeter of $\triangle$ is $x$ cm, then the area of $\Delta$ is $x$ square cm. Determine the lengths of the three sides of $\triangle$. | Let $\triangle$ have the lengths of its two legs as $a, b$, then
$$
\begin{array}{l}
\frac{1}{2} a b=a+b+\sqrt{a^{2}+b^{2}}, \\
4\left(a^{2}+b^{2}\right)=(2 a+2 b-a b)^{2}, \\
a^{2} b^{2}-4 a^{2} b-4 a b^{2}+8 a b=0, \\
a b-4 a-4 b+8=0, \\
(a-4)(b-4)=8 .
\end{array}
$$
Since $a, b$ are natural numbers, therefore
(1) wh... | 5,12,13 \text{ or } 6,8,10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,325 |
1. The students of Class (1), Grade 3 in a certain school plan to go mountain climbing on Sunday. They plan to set off at 8:30 am, and try to climb the farthest mountain on the map. After reaching the summit, they will hold a 1.5-hour cultural and entertainment activity, and must return to the base before 3 pm. If the ... | $-、 1 . D$.
Assuming the distance from the starting point to the top of the mountain is $x$ kilometers, then $\frac{x}{3.2}+\frac{x}{4.5}+1.5 \leqslant 6.5$.
Solving for $x$ gives $x \leqslant 9.3$. | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 708,326 |
2. Let the six-digit number $N=\overline{x 1527 y}$ be a multiple of 4, and $N$ leaves a remainder of 5 when divided by 11. Then, $x+y$ equals ( ).
(A) 8
(B) 9
(C) 10
(D) 11 | 2. B.
Since $N$ is a multiple of 4, $\therefore \overline{7 y}=70 x y$ is a multiple of 4,
$\therefore y=2$ or $y=6$.
If $y=2$, then $\overline{15278}$ is a multiple of 11,
$(x+5+7)-(1+2+3)=x+1$ is a multiple of 11,
no solution.
If $y=6$, then $\overline{x 15271}$ is a multiple of 11, i.e., $(x+5+7)-(1+2+1) \approx x+8... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,327 |
3. Let $a>0$, then the equation $\sqrt{a-x^{2}}=\sqrt{2}-|x|$ has unequal real roots. Therefore, the range of values for $a$ is ().
(A) $a>0$
(B) $0<a<1$
(C) $a=1$
(D) $a \geqslant 1$ | 3. D.
Square both sides of the equation, we get
$$
a-x^{2}=2+x^{2}-2 \sqrt{2}|x| \text {. }
$$
Rearrange and square again, we get $\left(a-2-2 x^{2}\right)^{2}=8 x^{2}$,
simplify to $\quad 4 x^{4}-4 a x^{2}+a^{2}-4 a+4=0$.
It is easy to see that, when $x \neq 0$, if $x$ is a root of the original equation, then $-x$ i... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 708,328 |
Example 8. Positive numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25, \\
\frac{y^{2}}{3}+z^{2}=9, \\
z^{2}+x z+x^{2}=16 .
\end{array}\right.
$$
Find the value of $x y+2 y z+3 x z$.
(18th All-Soviet Union Mathematical Olympiad) | Solve as shown in Figure 4, construct
a right-angled triangle with side lengths of 3,
4, and 5, and
make $\angle B P C=90^{\circ}, \angle A P C$
$=150^{\circ}$, then $\angle A P B=$
$120^{\circ}$. By the cosine rule, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{2}+\left(\frac{y}{\sqrt{3}}\right)^{2}-2 x \cdot ... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,329 |
4. Let $x \geqslant 0, y \geqslant 0, 2x + y = 6$. Then the maximum (minimum) value of $P = 4x^2 + 3xy + y^2 - 6x - 3y$ is ( ).
(A) has a maximum value of 18, no minimum value
(B) has no maximum value, has a minimum value of $\frac{27}{2}$
(C) has a maximum value of 18, a minimum value of $\frac{27}{2}$
(D) has neither... | 4. C.
Substitute $y=6-2 x(0 \leqslant x \leqslant 3)$ into $P$, we get
$$
P=2 x^{2}-6 x+18=2\left(x-\frac{3}{2}\right)^{2}+\frac{27}{2} \text {. }
$$
$\therefore$ When $x=\frac{3}{2}$, $P$ has the minimum value $\frac{27}{2}$. When $x=0$ or 3, $P$ has the maximum value 18. | null | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 708,330 |
5. A factory's timing clock takes 69 minutes for the minute hand to meet the hour hand once. If the hourly wage for workers is 4 yuan, and the overtime hourly wage for exceeding the specified time is 6 yuan, how much should be paid to a worker who completes the specified 8-hour work according to this clock?
(A) 34.3 yu... | 5. B.
$\because$ The clock should match the actual time for 8 hours
$$
8 \times \frac{69}{60+\frac{60}{11}}=\frac{253}{30} \text { (hours), }
$$
Therefore, the payment should be $6\left(\frac{253}{30}-8\right)+4 \times 8=34.6$ yuan. | 34.6 | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 708,331 |
6. As shown in the figure, $M$ is the midpoint of arc $C A B$, line segment $M P$ is perpendicular to chord $A B$ at $P$, and $A C=1 \mathrm{~cm}, P B=96 \mathrm{~cm}$, the length of $P A$ is:
(A) $77 \mathrm{~cm}$
(B) $96 \mathrm{~cm}$
(C) $67 \mathrm{~cm}$
(D) $115 \mathrm{~cm}$ | 6. A.
On $BP$, intercept $BD=AC$.
$$
\begin{array}{l}
\because \overparen{M C}=\widehat{M B}, \\
\therefore M B=M C, \\
\text { and } \angle B=\angle C,
\end{array}
$$
$\therefore \triangle M B D \cong \triangle M C A$. $C$
Thus, $M A=M D$.
Also, $M P \perp A D$,
Therefore, $A P=P D$.
So, $P A=P D=96-19=7(\mathrm{~cm}... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,332 |
1. Given real numbers $a, b, c$ satisfy $a+b+c=0, a b c>0$, and $x=\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}, y=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+$ $c\left(\frac{1}{a}+\frac{1}{b}\right)$. Then the algebraic expression $x^{18}-96 x y+y^{3}=$ | $$
=\sqrt{ } .-316
$$
Given $-b c>0$, and $a+b+c=0$, we know that among $a, b, c$, there must be two negative numbers. Without loss of generality, assume $a>0$. Hence $x=-1$.
$$
\begin{array}{l}
\text { ini } \begin{array}{l}
y=\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b} \\
=\frac{b+c}{a}+... | -316 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,333 |
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