Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Ajit Athle.

Click the figure below to see the complete problem 1021.

## Friday, June 13, 2014

### Geometry Problem 1021: Triangle, Cevian, Double Angle, Sides

Labels:
cevian,
double angle,
triangle

Subscribe to:
Post Comments (Atom)

Draw CH prenpedicular to AB.

ReplyDeleteLet AH = x.

By pythagorean theorem AC^2 - AH^2 = CB^2 - BH^2 ----> B^2 - x^2 = (b + c/2)^ - ( c- x)^2

From short develop finally we get AH = x = 3c/8 - b/2

Now draw isoceles triangle CDP (continue AP ....)

Notice that:

1) <A = <CDP + <CPD = 2*<APC

D is on AP.

DeleteYes , sure but the solution is right......?

DeleteI mean DH = PH = 3c/4 - x = 3c/4 - 3c/8 +b/2 = 3c/8 + b/2

Deletethen DA = DH - x = 3c/8 + b/2 - 3c/8 + b/2 = b

from here CA = DA = b ----> <CDP + <CPD = 2*<APC = <A

I understand the idea of your solution.

DeleteYour step AH = x = 3c/8 - b/2 is OK

but your step: "I mean DH = PH = 3c/4 - x = ..."

should be: DH = AH then triangles AHC and DHC are congruent (SAS) then CD = b then ....

Does this problem have a purely geometrical proof? I proved it using trigonometry.

ReplyDeleteApplying Stewart's theorem,(cp=x)

ReplyDeleteb^2.c/4 + (b+c/2)^2.3c/4=c(x^2 +3c/4.c/4)

=>b^2+(b+c/2)^2.3=4(x^2+3c^2/16)

=>b^2+3b^2+3bc+3c^2/4=4x^2 +3c^2/4

=>b^2+3/4.bc=x^2 =>x^2=b(b+3c/4)

,',In ΔAPC, x^2=b(b+3c/4). By triangle properties, <A=2<APC

See my synthetic solution at my page https://www.facebook.com/photo.php?fbid=784793084934933&set=a.784793394934902.1073741826.100002127454113&type=1&theater

ReplyDeletewith C as center, radius b, make a circle O, circle O intercept AB at E, intercept CB at F;

ReplyDeleteExtend BC to intercept circle O at G.

CA=CE, /_ CAB = /_CEA=2x

so BA*BE=BF*BG; BA=c,let BE=y, BF=1/2c, BG=2b+1/2c;

so BF= b+1/4c,

so PE=b,

so trangle CEP is isoceles,

so /_ CAB =2 /_CPA

Extend PA to X such that XA = b. Let Y be the foot of the perpendicular from C to AP. Let AY = p.

ReplyDeleteThen b^2 - p^2 = (b+c/2)^2 - (c-p)^2 from which p = 3c/8 - b/2 and PY = 3c/8+b/2

So XY = PY and hence XC = CP and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

Solution 2

ReplyDeleteMark Z on AP such that CZ = b

Then AZ = 2p = 3c/4 - b (from my previous proof) and so PZ = b = ZC

Hence < CZA = 2P = A

Sumith Peiris

Moratuwa

Sri Lanka