2018 State Mathcounts Sprint 30

Art of Problem Solving

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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2018 State Mathcounts Sprint 30

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ObiWanKenoblowin

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ObiWanKenoblowin

#1 h Mar 26, 2025, 4:18 PM

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Is there a non cord-bash way to solve this? Please let me know, thanks!

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ysn613

#2 h Mar 26, 2025, 4:34 PM

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Can you post the problem?

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ObiWanKenoblowin

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ObiWanKenoblowin

#3 h Mar 26, 2025, 4:44 PM

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Search it up( it includes a diagram and as a new user I can’t post images)

This post has been edited 1 time. Last edited by ObiWanKenoblowin, Mar 26, 2025, 4:44 PM

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#4 h Mar 26, 2025, 4:46 PM

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I'm utterly baffled from that problem

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#5 h Mar 26, 2025, 5:55 PM • 1 Y

Y by ObiWanKenoblowin

You can use mass points. The triangles $DPB$ and $APC$ intersect along the altitude from $P$ . Since $D'B'$ and $AC'$ are both in the plane, their intersection with the altitude is also in the plane. We can use triangle $DPB$ to find the mass at that intersection, like so:

$[asy] import graph; import olympiad; size(300); pair D = (0,0); pair B = (sqrt(2), 0); pair H = (B + D) / 2; pair P = H + (0, 1); pair Dp = (4 * D + P) / 5; pair Bp= (3 * B + P) / 4; pair I = intersectionpoint(Bp--Dp, P--H); draw(P--B--D--cycle); draw(Bp--Dp); draw(P--H); dot(Dp); dot(Bp); dot(H); dot(I); label("$3\ P\ 4$", P, N); label("$12\ B$", B, SE); label("$D\ 12$", D, SW); label("$D'\ 15$", Dp, NW); label("$16\ B'$", Bp, NE); label("$24$", H, S); label("$I\ \ 31$", I, N); [/asy]$

Then we can use triangle $APC$ to find the mass at $C'$ :

$[asy] import graph; import olympiad; size(300); pair A = (0,0); pair C = (sqrt(2), 0); pair H = (C + A) / 2; pair P = H + (0, 1); pair Cp= (12 * C + 7 * P) / 19; pair I = intersectionpoint(Cp--A, P--H); draw(P--C--A--cycle); draw(Cp--A); draw(P--H); dot(Cp); dot(H); dot(I); label("$P\ 7$", P, N); label("$12\ C$", C, SE); label("$A\ 12$", A, SW); label("$19\ C'$", Cp, NE); label("$24$", H, S); label("$I\ \ 31$", I, N); [/asy]$

This gives $\frac{C'C}{PC} = \frac{7}{19}.$

This post has been edited 4 times. Last edited by programjames1, Mar 26, 2025, 7:51 PM

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ObiWanKenoblowin

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ObiWanKenoblowin

#6 h Apr 2, 2025, 8:42 PM

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Thank you!

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