ข้อสอบ IWYMIC 2007
 

ประเภทบุคคล http://www.taimc2012.org/problem/200...Individual.pdf
ประเภททีม http://www.taimc2012.org/problem/2007-IWYMIC-Team.pdf
:please::please::please::please:

ประเภทบุคคล

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มาประเดิมข้อ 3 ให้ก่อนครับ


สามเหลี่ยม ABC มี E อยู่บนด้าน AC และ F อยู่บนด้าน AB, BE และ CF ตัดกันที่จุด D
ถ้าพื้นที่ของสามเหลี่ยม BDF, BCD และ CDE เท่ากับ 3, 7 และ 7 ตามลำดับแล้ว
พื้นที่สี่เหลี่ยม AEDF เป็นเท่าไร

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จากรูป ก
D แบ่งครึ่ง BE
สามเหลี่ยม DEF เท่ากับ 3

FD : DC = 3 : 7
ให้พื้นที่ AEF = x
พื้นที่สามเหลี่ยม ABC = x +20

รูป ข
สามเหลี่ยม ABE = x +6
ลาก AD
สามเหลี่ยม ABD = สามเหลี่ยม ADE = $\frac{1}{2}(x+6) = \frac{x}{2} +3$

สามเหลี่ยม$ \ AFD = \frac{x}{2} \ \ $

$ \frac{สามเหลี่ยม AFD }{สามเหลี่ยม ADC } = \frac{\frac{x}{2} }{ \frac{x}{2} +3 +7} = \frac{3}{7}$

x = 15

พื้นที่สี่เหลี่ยม AEDF = 15+3 = 18 ตารางหน่วย ซ.ต.พ.

ขอเป็นข้อ 11 นะครับ รีบไปหน่อยไม่ได้ละเอียดมากมายนะครับ

มีเฉลยคำตอบอย่างเดียว ไม่มีวิธีแสดงในหน้านี้ครับ Problem IWYMIC

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The smallest multiple of n is of course n. Denote by an the largest multiple of n not exceeding 101.
Then $A_n$=(n+$a_n$)/2.
Hence $A_2$=$A_3$=51, $A_4$=52, $A_5$=52.5 and $A_6$=51,
and the largest one is $A_5$.

2. It is a dark and stormy night. Four people must evacuate from an island to the mainland. The only link is a narrow bridge which allows passage of two people at a time. Moreover, the bridge must be illuminated, and the four people have only one lantern among them. After each passage to the mainland, if there are still people on the island, someone must bring the lantern back. Crossing the bridge individually, the four people take 2, 4, 8 and 16 minutes respectively. Crossing the bridge in pairs, the slower speed is used. What is the minimum time for the whole evacuation?
Solution Exactly five passages are required,
three pairs to the mainland and two individuals back to the island.
Let the fastest two people cross first.
One of them brings back the lantern.
Then the slowest two people cross,
and the fastest people on the mainland brings back the lantern,
The final passage is the same as the first.
The total time is 4+2+16+4+4=30 minutes.
To show that this is minimum,
note that the three passages in pairs take at least 16+4+4=24 minutes,
and the two passages individually take at least 4+2=6 minutes.

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Let the dimensions of the rectangle be x by y, with x≦y.
Then the number of soldiers on the outside is 2x+2y-4 while the number of those in the interior is (x-2)(y-2).
From xy-2x-2y+4=2x+2y-4,
we have (x-4)(y-4)=xy-4x-4y+16=8.
If x-4=2 and y-4=4, we obtain the original 6x8 rectangle.
If x-4=1 and y-4=8, we obtain the new 5x12 rectangle.
Thus the number of new soldiers is 5x12-6x8=12.

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Let a be the smallest of these integers.
Then a+(a+1)+(a+2)+?+(a+2007)=251x(2a+2007)x$2^2$.
In order for this to be a perfect square,
we must have 2a+2007=251$n^2$ for some positive integer n.
For n=1 or 2, a is negative.
For n=3, we have a=126
so that a+2007=2133 is the desired minimum value.

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In the first diagram, the ratio of the areas of the shaded triangle and one of the unshaded triangle is (5－3):3
so that $S_A$ is one quarter of the area of the whole triangle.

In the second diagram, the ratio of the areas of the shaded triangle and one of the unshaded triangle is (5－4):4
so that $S_B$ is one ninth of the area of the whole triangle.

Now 1/4+1/9=13/36.
Hence the area of the whole triangle is (36/13)39=108 square centimetres.

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Note that $3^{1024}$-1=($3^{512}$+1)($3^{256}$+1)($3^{128}$+1)?(3+1)(3-1).
All 11 factors are even, and 3+1 is a multiple of 4.
Clearly 3-1 is not divisible by 4.
We claim that neither is any of the other 9.
When the square of an odd number is divided by 4,
the remainder is always 1.
Adding 1 makes the remainder 2, justifying the claim.
Hence the maximum value of n is 12.

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We may assume that the sides of lengths 1 and 8 are adjacent sides of the quadrilateral,
as otherwise we can flip over the shaded triangle in the first diagram.
Now the quadrilateral may be divided into two triangles as shown in the second diagram.
In each triangle, two sides have fixed length.

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Hence its area is maximum if these two sides are perpendicular to each other.
Since $1^2$+$8^2$=$4^2$+$7^2$,
both maxima can be achieved simultaneously.
In that case, the area of the unshaded triangle is 4
and the area of the shaded triangle is 14.
Hence the maximum area of the quadrilateral is 18.

Answer : $\frac{7}{8}$

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ข้อ 4 ประเภททีมนะครับ ... อาจจะไม่เข้าใจเเต่ก็พยายามเขียนเต็มที่เเล้วอะครับ :unsure:

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