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

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

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## Sunday, November 13, 2016

### Geometry Problem 1287 Isosceles Triangle, 40-100-40 Degrees, Congruence, Area, Metric Relations, Measurement

Labels:
40 degrees,
40-100-40,
area,
congruence,
isosceles,
measurement,
metric relations,
triangle

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Problem 1287

ReplyDeleteIn extension of the CB passes point D such that AD=AC=b.Then the triangles ABC and CAD are similar.So a/b=b/CD or CD=b^2/a,BD=CD-CB=b^2/a-a. But <BAD=100-40=60,then BD^2=

a^2+b^2-ab =(b^2/a-a)^2 so a^3+b^3=3a^2b.

Draw BK perpendicular in AC then BK^2=a^2-b^2/4.

Let area triangle ABC is (ABC)=(AC.BK)/2 then substituting the BKand using a^3+b^3=3a^2b we have the effect.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

From A, draw a line at 20 degrees to AC and length = a and form an equilateral triangle ABD.

ReplyDeleteLet BD intersect AC at E.

If you observe the triangle BEC is similar to ABC

Hence BE=EC=a^2/b

Therefore AE=b-a^2/b=b^2-a^2/b

=> AE^2=(b^2-a^2/b)^2 --------------(1)

Now consider the triangle ABE where m(ABE)=60

AE^2=AB^2+BE^2-2.AB.BE.COS60

=> AE^2=a^2+a^4/b^2-a^3/b---------------(2)

From (1) & (2)

(b^2-a^2/b)^2=a^2+a^4/b^2-a^3/b

=> a^3+b^3=3a^2b

Let area triangle ABC is (ABC) then (ABC)^2=(b^2.BK^2)/4=b^2(4a^2-b^2)/16. And (ABC)^2=3/144.b(4a^3+b^3).Let b^2(4a^2-b^2)/16=3/144.b(4a^3+b^3) or

ReplyDelete(4a^3+b^3)/3=b(4a^2-b^2) or a^3+b^3=3a^2b which applies.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE