## Saturday, October 27, 2012

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

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

Click the figure below to see the complete problem 819. 1. Triangle ABC is isosceles, AC = 2ABcos40
Using sine rule on triangle ACD,
AC/sinx = CD/sin30
2ABcos40/sinx = CD/sin30
cos40 = sinx
x = 50

1. why x not= 130?
I believe x=50 or 130.

2. In triangle ABC:
AC = 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

1. Why x=50 only?
I believe x=50 or 130.

3. 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).

1. Antonio, did you judge the answer according to figure?

2. To 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.
Thanks.

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

5. Let E be the circumcenter of ΔACD.

Then
∠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°

6. draw the altitudes BM of triangle ABC,CH of triangle ACD
CM=AC/2 (ABC is isoscele)
CH=AC/2 (triangle 30-60-90)
the right triangles BMC and CHD are congruent
x=50

7. Prof Radu Ion,Sc.Gim.Bozioru,BuzauOctober 29, 2012 at 10:08 AM

Costruind simetricul punctului C fata de dreapta AD si notandu-lcu P obtinem :
ΔACP echilateral si ΔACP=ΔPCD(LLL);∠CDP = ∠ABC = 100°,AD mediatoarea seg AD =>AD bisectoarea ∠CDP => ∠CDA = 50°

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

9. 10. Let BA=BC=CD=a and AC=d
Using 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.

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

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

13. En la figura adjunta resuelvo el problema sin usar CD...:D
http://www.subirimagenes.com/otros-ahorasi-8277920.html

By Tony García.

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

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

Sumith Peiris
Moratuwa
Sri Lanka

16. 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.

Sumith Peiris
Moratuwa
Sri Lanka

17. 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.

Best regards

18. Draw BE as perpendicular to AC and CF perpendicular to AD. Join EF.
BE 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°