olymmath

TR-MO-R1 2013 Problem 9

    ABCABC üçgeninde ∣AB∣=18|AB| = 18, ∣AC∣=24|AC| = 24 ve m(BAC^)=150∘m(\widehat{BAC}) = 150^\circ dir. DD noktası [AB][AB], EE noktası [AC][AC] ve FF noktası [BC][BC] kenarları üstünde olmak üzere, ∣BD∣=6|BD| = 6, ∣CE∣=8|CE| = 8 ve ∣CF∣=2∣BF∣|CF| = 2|BF| dir. ABCABC üçgeninin diklik merkezi HH noktasının DD, EE ve FF noktalarına göre simetrikleri sırasıyla, H1H_1, H2H_2 ve H3H_3 noktaları ise, H1H2H3H_1H_2H_3 üçgeninin alanı nedir?

    a) 70 b) 72 c) 84 d) 96 e) 108

    Official text Turkish Mathematical Olympiad, first round 2013 official paper, page 4. Quoted with source.

    A model read this text from the scan of the paper by OCR, with vision/claude@2.1.318 (claude-opus-5-5). No person has checked it against the paper yet.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator tr-mo-r1/2013/p9 · Tag QD41M5XRZV · Edit this page on GitHub · Keys: j next, k previous