I had a math comp. 3 days ago. And I can't do these problems. Stuck a lot. :<

1.) Triangle ABC given that AB = AC. Draw a line from B perpendicular to AC at point D. Draw another line from D perpendicular to BC at E. If BC = AB + AD, prove that BE = CD.

2.) Given a number \(x\) with more than 1 digit. If we write it twice (such as x = 137, we write it 137137), we'll get a number that is divisible by \(x^{2}\). Prove that the 2 first digits are 14......

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## Comments

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TopNewestI just figured no. 2 out. XD

Given that \(x^{2} | \overline{xx} \rightarrow x^{2} | x(10^{n}+1) \rightarrow x | (10^{n} + 1)\).

Factor \(10^{n} + 1\) as \(a\times b\) where \(a, b\) are positive integers.

Since we know that both of them are odd numbers, but not divisible by 3 and 5 since \(3, 5 \not| (10^{n} + 1)\).

So \(a, b\) must be divisible by 7.

When we divide \(10^{n} + 1\) by 7, it always starts with 1, 4 for \(n \geq 2\). Hence, proven. ~~~

I always hate myself when I can't do it during the test but after the test. >:(

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Let angle \(ACB\)'s size be \(x\).

Then angle \(EDC\)'s and angle \(EDB\)'s size will be \(90-x\) and \(x\), respectively.

From the properties, we will get that

\(\bigtriangleup BDE \sim \bigtriangleup BCD\)

Thus, \(\frac{BE}{BD} = \frac{BD}{BC}\)

\(BE = \frac{BD^{2}}{BC} = \frac{AB^{2} - AD^{2}}{AB + BD}\) (Pythagorean Theorem)\( = AB - AD = AC - AD \)(Isosceles Triangle)\(= DC\) \(Q.E.D.\) ซ.ต.พ.

พิสูจน์สองข้อนี้ผมแป้กในห้องสอบทั้งคู่ครับ ไม่ทราบว่าได้เหรียญอะไรอ่อครับ

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ได้แค่เหรียญทองแดงเองคับ เสียไป 2 ข้อนี้ฟรีๆ เหมือนเส้นผมบังภูเขา T__T PS: ข้อแรกเพิ่งคิดได้ตอนนั่งคิดเล่นๆ ใช้เวลาแค่ 5 นาทีเสร็จ ==''

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ผมก็ได้แค่เหรียญทองแดงครับ เจอเรขานี้ Let it go แต่ข้อ Number Theory แสดงไปซุยๆครับแค่สองบรรทัดแรก 555

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Hi Samuraiwarm!

Sorry, but I wasn't able to solve any of them. But could you please do me a favour?

Please send all the questions to me, I want to give all a try. My email address is sgsuper@yahoo.com, thanks.

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