Maths Olympiad Prep

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, 2017

Algebra Difficulty 2.1 Junior Prove it Canada

By finding a common denominator, we see that 13\dfrac{1}{3} is greater than 17\dfrac{1}{7} because 721>321\dfrac{7}{21}>\dfrac{3}{21}.

Similarly, we see that 13\dfrac{1}{3} is less than 12\dfrac{1}{2} because 26<36\dfrac{2}{6}<\dfrac{3}{6}.

Determine the integer nn so that n40\dfrac{n}{40} is greater than 15\dfrac{1}{5} and less than 14\dfrac{1}{4}.
Determine all possible integers mm so that m8\dfrac{m}{8} is greater than 13\dfrac{1}{3} and m+18\dfrac{m+1}{8} is less than 23\dfrac{2}{3}.
Fiona calculates her win ratio by dividing the number of games that she has won by the total number of games that she has played. At the start of a weekend, Fiona has played 30 games, has ww wins, and her win ratio is greater than 0.5. During the weekend, she plays five games and wins three of these games. At the end of the weekend, Fiona’s win ratio is less than 0.7. Determine all possible values of ww.

Solution

Chords DEDE and FGFG intersect at XX, and so (DX)(EX)=(FX)(GX)(DX)(EX)=(FX)(GX) or (DX)(8)=(6)(4)(DX)(8)=(6)(4).

Solving this equation for DXDX, we get DX=(6)(4)8=248=3DX=\dfrac{(6)(4)}{8}=\dfrac{24}{8}=3.

The length of DXDX is 3.
Chords JKJK and LMLM intersect at XX, and so (JX)(KX)=(LX)(MX)(JX)(KX)=(LX)(MX) or(8y)(10)=(16)(y+9)(8y)(10)=(16)(y+9).

Solving this equation, we get 80y=16y+14480y=16y+144 or 64y=14464y=144, and so y=14464=94y=\dfrac{144}{64}=\dfrac{9}{4}.
Chords PQPQ and STST intersect at UU, and so (PU)(QU)=(SU)(TU)(PU)(QU)=(SU)(TU).

Since TU=TV+UVTU=TV+UV, then TU=6+nTU=6+n.

Substituting values into (PU)(QU)=(SU)(TU)(PU)(QU)=(SU)(TU), we get (m)(5)=(3)(6+n)(m)(5)=(3)(6+n), and so 5m=18+3n5m=18+3n.

Chords PRPR and STST intersect at VV, and so (PV)(RV)=(TV)(SV)(PV)(RV)=(TV)(SV).

Since SV=SU+UVSV=SU+UV, then SV=3+nSV=3+n.

Substituting values into (PV)(RV)=(TV)(SV)(PV)(RV)=(TV)(SV), we get (n)(8)=(6)(3+n)(n)(8)=(6)(3+n), and so 8n=18+6n8n=18+6n or 2n=182n=18 or n=9n=9.

Substituting n=9n=9 into 5m=18+3n5m=18+3n, we get 5m=18+3(9)5m=18+3(9) or 5m=455m=45, and so m=9m=9.

Therefore, m=9m=9 and n=9n=9.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.