Geometry Problem

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

Click the figure below to see the complete problem 819.

## Saturday, October 27, 2012

### Problem 819: Quadrilateral, Triangle, Angles, 30 degrees

Labels:
30 degrees,
angle,
quadrilateral,
triangle

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Triangle ABC is isosceles, AC = 2ABcos40

ReplyDeleteUsing sine rule on triangle ACD,

AC/sinx = CD/sin30

2ABcos40/sinx = CD/sin30

cos40 = sinx

x = 50

why x not= 130?

DeleteI believe x=50 or 130.

In triangle ABC:

ReplyDeleteAC = 2 AB cos(40) ---- (1)

In triangle ACD:

AC = CD sin(x)/sin(30) = 2 CD sin(x) ---- (2)

AB = CD;

Then by (1) and (2):

sin(x) = cos(40) = sin(50)

x = 50

Why x=50 only?

DeleteI believe x=50 or 130.

To Anonymous: according to the figure angle x is acute, therefore x=50. If D' is on AD so that AB=BC=CD', then angle x is obtuse (x=130).

ReplyDeleteAntonio, did you judge the answer according to figure?

DeleteTo Anonymous: I think the problem has two solutions. First, according to the figure, when the angle x is acute and the second, when the angle x is obtuse.

DeleteThanks.

Problem 819: Try to use elementary geometry (Euclid's Elements).

ReplyDeleteLet E be the circumcenter of ΔACD.

ReplyDeleteThen

∠CED = 2×∠CAD = 60°, CD = CE

⇒ ΔCDE is equilateral

⇒ AB = BC = CD = DE = CE = AE

⇒ ABCE is a rhombus

⇒ ∠AEC = ∠ABC = 100°

⇒ ∠ADC = 1/2×∠AEC = 50°

draw the altitudes BM of triangle ABC,CH of triangle ACD

ReplyDeleteCM=AC/2 (ABC is isoscele)

CH=AC/2 (triangle 30-60-90)

the right triangles BMC and CHD are congruent

x=50

Costruind simetricul punctului C fata de dreapta AD si notandu-lcu P obtinem :

ReplyDeleteΔACP echilateral si ΔACP=ΔPCD(LLL);∠CDP = ∠ABC = 100°,AD mediatoarea seg AD =>AD bisectoarea ∠CDP => ∠CDA = 50°

Prof Radu Ion,Sc.Gim.Bozioru,Buzau

Draw BE perpendicular to AC and CF perpendicular to AD. Join EF.

ReplyDeleteBE bisects AC, since AB = BC

E being the midpoint of the hypotenuse AC of the right triangle AFC,

EA = EC = EF and consequently ΔEFC is equilateral.

Now consider the right triangles BEC and DFC.

By Pythagoras, DF^2 = CD^2 - CF^2 = BC^2 - CE^2 = BE*2.

So DF = BE and consequently the right triangles CDF and CBE are congruent.

Follows x = ∠FBC.

Note that ΔABC is isosceles (with base angles FCB, FAB each 40°),

Hence x = ∠FBC = 90° - ∠FCB = 90° - 40° = 50°

x=50deg if AD>AC; or x=130deg IF AD<AC.

ReplyDeleteLet BA=BC=CD=a and AC=d

ReplyDeleteUsing sine rule,

In triangle ABC, a/sin 40=d/sin 100…(1)

In triangle ACD, a/sin 30=d/sin x…(2)

Using (1),(2), we get, sin 100/sin40=sin x/sin 30

sin 80/sin40=sin x/sin 30

2(sin 40)(cos 40)/sin40=sin x/(1/2)

cos 40=sin x

sin (90-40)=sin x

sin 50=sin x

50=x.

Let E be the point on AD such that AB= BE.

ReplyDeleteang. AEB= 700 (∆ ABE is isosceles)

ang ABC= 1000= ang. ABE+ ang. EBC → ang EBC= 600

Since BE=BC (by our construction) so,

ang. BEC= ang. ECB = 1800-600/2= 600

This implies ∆ BEC is equilateral. → CD= CE.

so ang. X= ang. CED = 1800-(700+600) = 500

□

Sorry for the extra zeros in the solution, they all refers degrees.

ReplyDeleteThank You.

En la figura adjunta resuelvo el problema sin usar CD...:D

ReplyDeletehttp://www.subirimagenes.com/otros-ahorasi-8277920.html

By Tony García.

http://www.mathematica.gr/forum/viewtopic.php?f=20&t=32307&p=149499

ReplyDeleteDraw altitudes BH and CG. Then by 30-60-90 Triangle and congruence x=50

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Another proof. Find E on AD such that AB = BE. Then < EBC = 60 and Tr. EBC is equilateral, Tr. ECD is isoceles and x = 50.

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Let D' on AD so that CD'=CD and <AD'C is obtuse and let B' be the circumcenter of tr. AD'C. Clearly tr CB'D' is equilateral and CB'=AB'=CD'=CB=AB, so B'=B, thus <AD'C=180-m(<ABC)/2=130 degs. and, as constructed, <ADC=<CD'D=180-m(<AD'C)=50 degs.

ReplyDeleteBest regards