HMMT 二月 2007 · 冲刺赛 · 第 30 题
HMMT February 2007 — Guts Round — Problem 30
题目详情
- [ 15 ] ABCD is a cyclic quadrilateral in which AB = 3 , BC = 5 , CD = 6, and AD = 10. M , I , and Tare the feet of the perpendiculars from D to lines AB, AC , and BC respectively. Determine the valueof M I/IT .
10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUNDth
3 2
英文原题
[ 15 ] ABCD is a cyclic quadrilateral in which AB = 3 , BC = 5 , CD = 6, and AD = 10. M , I , and T
are the feet of the perpendiculars from D to lines AB, AC , and BC respectively. Determine the value
of M I/IT .
4
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10 th HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND
解析
英文解析
- [ 15 ] ABCD is a cyclic quadrilateral in which AB = 3 , BC = 5 , CD = 6, and AD = 10. M , I , and Tare the feet of the perpendiculars from D to lines AB, AC , and BC respectively. Determine the valueof M I/IT .
Answer: . Quadrilaterals AM ID and DICT are cyclic, having right angles ∠ AM D, ∠ AID , and 259
∠ CID, ∠ CT D respectively. We see that M , I , and T are collinear. For, m ∠ M ID = π − m ∠ DAM =9
π − m ∠ DAB = m ∠ BCD = π − m ∠ DCT = π − m ∠ DIT . Therefore, Menelaus’ theorem applied totriangle M T B and line ICA gives
M I T C BA
· · = 1
IT CB AM
∼ ∼
On the other hand, triangle ADM is similar to triangle CDT since ∠ AM D ∠ CT D and ∠ DAM = =
∠ DCT and thus AM/CT = AD/CD . It follows that
M I BC · AM BC · AD 5 · 10 25 = = = =
IT AB · CT AB · CD 3 · 6 9
Remarks. The line M IT , constructed in this problem by taking perpendiculars from a point on thecircumcircle of ABC , is known as the Simson line . It is often helpful for us to use directed angles whileangle chasing to avoid supplementary configuration issues, such as those arising while establishing thecollinearity of M, I , and T .
10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUNDth
3 2